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提交 fb246063 编写于 作者: J jjb
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6804124: Replace "modified mergesort" in java.util.Arrays.sort with timsort 

Summary: Easy port of timsort from android
Reviewed-by: martin
上级 ba4e2b67
隐藏空白更改
内联 并排
Showing with 2389 addition and 93 deletion
make/java/java/FILES_java.gmk
浏览文件 @ fb246063
...@@ -250,6 +250,8 @@ JAVA_JAVA_java = \ ...@@ -250,6 +250,8 @@ JAVA_JAVA_java = \
java/util/IdentityHashMap.java \ java/util/IdentityHashMap.java \
java/util/EnumMap.java \ java/util/EnumMap.java \
java/util/Arrays.java \ java/util/Arrays.java \
java/util/TimSort.java \
java/util/ComparableTimSort.java \
java/util/ConcurrentModificationException.java \ java/util/ConcurrentModificationException.java \
java/util/ServiceLoader.java \ java/util/ServiceLoader.java \
java/util/ServiceConfigurationError.java \ java/util/ServiceConfigurationError.java \
......
src/share/classes/java/util/Arrays.java
浏览文件 @ fb246063
...@@ -1065,29 +1065,103 @@ public class Arrays { ...@@ -1065,29 +1065,103 @@ public class Arrays {
(x[b] > x[c] ? b : x[a] > x[c] ? c : a)); (x[b] > x[c] ? b : x[a] > x[c] ? c : a));
} }
/** /**
* Sorts the specified array of objects into ascending order, according to * Old merge sort implementation can be selected (for
* the {@linkplain Comparable natural ordering} * compatibility with broken comparators) using a system property.
* of its elements. All elements in the array * Cannot be a static boolean in the enclosing class due to
* must implement the {@link Comparable} interface. Furthermore, all * circular dependencies. To be removed in a future release.
* elements in the array must be <i>mutually comparable</i> (that is, */
* <tt>e1.compareTo(e2)</tt> must not throw a <tt>ClassCastException</tt> static final class LegacyMergeSort {
* for any elements <tt>e1</tt> and <tt>e2</tt> in the array).<p> private static final boolean userRequested =
java.security.AccessController.doPrivileged(
new sun.security.action.GetBooleanAction(
"java.util.Arrays.useLegacyMergeSort")).booleanValue();
}
/*
* If this platform has an optimizing VM, check whether ComparableTimSort
* offers any performance benefit over TimSort in conjunction with a
* comparator that returns:
* {@code ((Comparable)first).compareTo(Second)}.
* If not, you are better off deleting ComparableTimSort to
* eliminate the code duplication. In other words, the commented
* out code below is the preferable implementation for sorting
* arrays of Comparables if it offers sufficient performance.
*/
// /**
// * A comparator that implements the natural ordering of a group of
// * mutually comparable elements. Using this comparator saves us
// * from duplicating most of the code in this file (one version for
// * Comparables, one for explicit Comparators).
// */
// private static final Comparator<Object> NATURAL_ORDER =
// new Comparator<Object>() {
// @SuppressWarnings("unchecked")
// public int compare(Object first, Object second) {
// return ((Comparable<Object>)first).compareTo(second);
// }
// };
//
// public static void sort(Object[] a) {
// sort(a, 0, a.length, NATURAL_ORDER);
// }
//
// public static void sort(Object[] a, int fromIndex, int toIndex) {
// sort(a, fromIndex, toIndex, NATURAL_ORDER);
// }
/**
* Sorts the specified array of objects into ascending order, according
* to the {@linkplain Comparable natural ordering} of its elements.
* All elements in the array must implement the {@link Comparable}
* interface. Furthermore, all elements in the array must be
* <i>mutually comparable</i> (that is, {@code e1.compareTo(e2)} must
* not throw a {@code ClassCastException} for any elements {@code e1}
* and {@code e2} in the array).
*
* <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* not be reordered as a result of the sort.
*
* <p>Implementation note: This implementation is a stable, adaptive,
* iterative mergesort that requires far fewer than n lg(n) comparisons
* when the input array is partially sorted, while offering the
* performance of a traditional mergesort when the input array is
* randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
* *
* This sort is guaranteed to be <i>stable</i>: equal elements will * <p>The implementation takes equal advantage of ascending and
* not be reordered as a result of the sort.<p> * descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
* *
* The sorting algorithm is a modified mergesort (in which the merge is * <p>The implementation was adapted from Tim Peters's list sort for Python
* omitted if the highest element in the low sublist is less than the * (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* lowest element in the high sublist). This algorithm offers guaranteed * TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* n*log(n) performance. * Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
* *
* @param a the array to be sorted * @param a the array to be sorted
* @throws ClassCastException if the array contains elements that are not * @throws ClassCastException if the array contains elements that are not
* <i>mutually comparable</i> (for example, strings and integers). * <i>mutually comparable</i> (for example, strings and integers)
* @throws IllegalArgumentException (optional) if the natural
* ordering of the array elements is found to violate the
* {@link Comparable} contract
*/ */
public static void sort(Object[] a) { public static void sort(Object[] a) {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a);
else
ComparableTimSort.sort(a);
}
/** To be removed in a future release. */
private static void legacyMergeSort(Object[] a) {
Object[] aux = a.clone(); Object[] aux = a.clone();
mergeSort(aux, a, 0, a.length, 0); mergeSort(aux, a, 0, a.length, 0);
} }
...@@ -1097,34 +1171,63 @@ public class Arrays { ...@@ -1097,34 +1171,63 @@ public class Arrays {
* ascending order, according to the * ascending order, according to the
* {@linkplain Comparable natural ordering} of its * {@linkplain Comparable natural ordering} of its
* elements. The range to be sorted extends from index * elements. The range to be sorted extends from index
* <tt>fromIndex</tt>, inclusive, to index <tt>toIndex</tt>, exclusive. * {@code fromIndex}, inclusive, to index {@code toIndex}, exclusive.
* (If <tt>fromIndex==toIndex</tt>, the range to be sorted is empty.) All * (If {@code fromIndex==toIndex}, the range to be sorted is empty.) All
* elements in this range must implement the {@link Comparable} * elements in this range must implement the {@link Comparable}
* interface. Furthermore, all elements in this range must be <i>mutually * interface. Furthermore, all elements in this range must be <i>mutually
* comparable</i> (that is, <tt>e1.compareTo(e2)</tt> must not throw a * comparable</i> (that is, {@code e1.compareTo(e2)} must not throw a
* <tt>ClassCastException</tt> for any elements <tt>e1</tt> and * {@code ClassCastException} for any elements {@code e1} and
* <tt>e2</tt> in the array).<p> * {@code e2} in the array).
*
* <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* not be reordered as a result of the sort.
*
* <p>Implementation note: This implementation is a stable, adaptive,
* iterative mergesort that requires far fewer than n lg(n) comparisons
* when the input array is partially sorted, while offering the
* performance of a traditional mergesort when the input array is
* randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
* *
* This sort is guaranteed to be <i>stable</i>: equal elements will * <p>The implementation takes equal advantage of ascending and
* not be reordered as a result of the sort.<p> * descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
* *
* The sorting algorithm is a modified mergesort (in which the merge is * <p>The implementation was adapted from Tim Peters's list sort for Python
* omitted if the highest element in the low sublist is less than the * (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* lowest element in the high sublist). This algorithm offers guaranteed * TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* n*log(n) performance. * Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
* *
* @param a the array to be sorted * @param a the array to be sorted
* @param fromIndex the index of the first element (inclusive) to be * @param fromIndex the index of the first element (inclusive) to be
* sorted * sorted
* @param toIndex the index of the last element (exclusive) to be sorted * @param toIndex the index of the last element (exclusive) to be sorted
* @throws IllegalArgumentException if <tt>fromIndex &gt; toIndex</tt> * @throws IllegalArgumentException if {@code fromIndex > toIndex} or
* @throws ArrayIndexOutOfBoundsException if <tt>fromIndex &lt; 0</tt> or * (optional) if the natural ordering of the array elements is
* <tt>toIndex &gt; a.length</tt> * found to violate the {@link Comparable} contract
* @throws ClassCastException if the array contains elements that are * @throws ArrayIndexOutOfBoundsException if {@code fromIndex < 0} or
* not <i>mutually comparable</i> (for example, strings and * {@code toIndex > a.length}
* integers). * @throws ClassCastException if the array contains elements that are
* not <i>mutually comparable</i> (for example, strings and
* integers).
*/ */
public static void sort(Object[] a, int fromIndex, int toIndex) { public static void sort(Object[] a, int fromIndex, int toIndex) {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, fromIndex, toIndex);
else
ComparableTimSort.sort(a, fromIndex, toIndex);
}
/** To be removed in a future release. */
private static void legacyMergeSort(Object[] a,
int fromIndex, int toIndex) {
rangeCheck(a.length, fromIndex, toIndex); rangeCheck(a.length, fromIndex, toIndex);
Object[] aux = copyOfRange(a, fromIndex, toIndex); Object[] aux = copyOfRange(a, fromIndex, toIndex);
mergeSort(aux, a, fromIndex, toIndex, -fromIndex); mergeSort(aux, a, fromIndex, toIndex, -fromIndex);
...@@ -1133,6 +1236,7 @@ public class Arrays { ...@@ -1133,6 +1236,7 @@ public class Arrays {
/** /**
* Tuning parameter: list size at or below which insertion sort will be * Tuning parameter: list size at or below which insertion sort will be
* used in preference to mergesort or quicksort. * used in preference to mergesort or quicksort.
* To be removed in a future release.
*/ */
private static final int INSERTIONSORT_THRESHOLD = 7; private static final int INSERTIONSORT_THRESHOLD = 7;
...@@ -1142,6 +1246,7 @@ public class Arrays { ...@@ -1142,6 +1246,7 @@ public class Arrays {
* low is the index in dest to start sorting * low is the index in dest to start sorting
* high is the end index in dest to end sorting * high is the end index in dest to end sorting
* off is the offset to generate corresponding low, high in src * off is the offset to generate corresponding low, high in src
* To be removed in a future release.
*/ */
private static void mergeSort(Object[] src, private static void mergeSort(Object[] src,
Object[] dest, Object[] dest,
...@@ -1197,25 +1302,53 @@ public class Arrays { ...@@ -1197,25 +1302,53 @@ public class Arrays {
* Sorts the specified array of objects according to the order induced by * Sorts the specified array of objects according to the order induced by
* the specified comparator. All elements in the array must be * the specified comparator. All elements in the array must be
* <i>mutually comparable</i> by the specified comparator (that is, * <i>mutually comparable</i> by the specified comparator (that is,
* <tt>c.compare(e1, e2)</tt> must not throw a <tt>ClassCastException</tt> * {@code c.compare(e1, e2)} must not throw a {@code ClassCastException}
* for any elements <tt>e1</tt> and <tt>e2</tt> in the array).<p> * for any elements {@code e1} and {@code e2} in the array).
*
* <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* not be reordered as a result of the sort.
*
* <p>Implementation note: This implementation is a stable, adaptive,
* iterative mergesort that requires far fewer than n lg(n) comparisons
* when the input array is partially sorted, while offering the
* performance of a traditional mergesort when the input array is
* randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
* *
* This sort is guaranteed to be <i>stable</i>: equal elements will * <p>The implementation takes equal advantage of ascending and
* not be reordered as a result of the sort.<p> * descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
* *
* The sorting algorithm is a modified mergesort (in which the merge is * <p>The implementation was adapted from Tim Peters's list sort for Python
* omitted if the highest element in the low sublist is less than the * (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* lowest element in the high sublist). This algorithm offers guaranteed * TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* n*log(n) performance. * Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
* *
* @param a the array to be sorted * @param a the array to be sorted
* @param c the comparator to determine the order of the array. A * @param c the comparator to determine the order of the array. A
* <tt>null</tt> value indicates that the elements' * {@code null} value indicates that the elements'
* {@linkplain Comparable natural ordering} should be used. * {@linkplain Comparable natural ordering} should be used.
* @throws ClassCastException if the array contains elements that are * @throws ClassCastException if the array contains elements that are
* not <i>mutually comparable</i> using the specified comparator. * not <i>mutually comparable</i> using the specified comparator
* @throws IllegalArgumentException (optional) if the comparator is
* found to violate the {@link Comparator} contract
*/ */
public static <T> void sort(T[] a, Comparator<? super T> c) { public static <T> void sort(T[] a, Comparator<? super T> c) {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, c);
else
TimSort.sort(a, c);
}
/** To be removed in a future release. */
private static <T> void legacyMergeSort(T[] a, Comparator<? super T> c) {
T[] aux = a.clone(); T[] aux = a.clone();
if (c==null) if (c==null)
mergeSort(aux, a, 0, a.length, 0); mergeSort(aux, a, 0, a.length, 0);
...@@ -1226,36 +1359,65 @@ public class Arrays { ...@@ -1226,36 +1359,65 @@ public class Arrays {
/** /**
* Sorts the specified range of the specified array of objects according * Sorts the specified range of the specified array of objects according
* to the order induced by the specified comparator. The range to be * to the order induced by the specified comparator. The range to be
* sorted extends from index <tt>fromIndex</tt>, inclusive, to index * sorted extends from index {@code fromIndex}, inclusive, to index
* <tt>toIndex</tt>, exclusive. (If <tt>fromIndex==toIndex</tt>, the * {@code toIndex}, exclusive. (If {@code fromIndex==toIndex}, the
* range to be sorted is empty.) All elements in the range must be * range to be sorted is empty.) All elements in the range must be
* <i>mutually comparable</i> by the specified comparator (that is, * <i>mutually comparable</i> by the specified comparator (that is,
* <tt>c.compare(e1, e2)</tt> must not throw a <tt>ClassCastException</tt> * {@code c.compare(e1, e2)} must not throw a {@code ClassCastException}
* for any elements <tt>e1</tt> and <tt>e2</tt> in the range).<p> * for any elements {@code e1} and {@code e2} in the range).
*
* <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* not be reordered as a result of the sort.
*
* <p>Implementation note: This implementation is a stable, adaptive,
* iterative mergesort that requires far fewer than n lg(n) comparisons
* when the input array is partially sorted, while offering the
* performance of a traditional mergesort when the input array is
* randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
* *
* This sort is guaranteed to be <i>stable</i>: equal elements will * <p>The implementation takes equal advantage of ascending and
* not be reordered as a result of the sort.<p> * descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
* *
* The sorting algorithm is a modified mergesort (in which the merge is * <p>The implementation was adapted from Tim Peters's list sort for Python
* omitted if the highest element in the low sublist is less than the * (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* lowest element in the high sublist). This algorithm offers guaranteed * TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* n*log(n) performance. * Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
* *
* @param a the array to be sorted * @param a the array to be sorted
* @param fromIndex the index of the first element (inclusive) to be * @param fromIndex the index of the first element (inclusive) to be
* sorted * sorted
* @param toIndex the index of the last element (exclusive) to be sorted * @param toIndex the index of the last element (exclusive) to be sorted
* @param c the comparator to determine the order of the array. A * @param c the comparator to determine the order of the array. A
* <tt>null</tt> value indicates that the elements' * {@code null} value indicates that the elements'
* {@linkplain Comparable natural ordering} should be used. * {@linkplain Comparable natural ordering} should be used.
* @throws ClassCastException if the array contains elements that are not * @throws ClassCastException if the array contains elements that are not
* <i>mutually comparable</i> using the specified comparator. * <i>mutually comparable</i> using the specified comparator.
* @throws IllegalArgumentException if <tt>fromIndex &gt; toIndex</tt> * @throws IllegalArgumentException if {@code fromIndex > toIndex} or
* @throws ArrayIndexOutOfBoundsException if <tt>fromIndex &lt; 0</tt> or * (optional) if the comparator is found to violate the
* <tt>toIndex &gt; a.length</tt> * {@link Comparator} contract
* @throws ArrayIndexOutOfBoundsException if {@code fromIndex < 0} or
* {@code toIndex > a.length}
*/ */
public static <T> void sort(T[] a, int fromIndex, int toIndex, public static <T> void sort(T[] a, int fromIndex, int toIndex,
Comparator<? super T> c) { Comparator<? super T> c) {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, fromIndex, toIndex, c);
else
TimSort.sort(a, fromIndex, toIndex, c);
}
/** To be removed in a future release. */
private static <T> void legacyMergeSort(T[] a, int fromIndex, int toIndex,
Comparator<? super T> c) {
rangeCheck(a.length, fromIndex, toIndex); rangeCheck(a.length, fromIndex, toIndex);
T[] aux = copyOfRange(a, fromIndex, toIndex); T[] aux = copyOfRange(a, fromIndex, toIndex);
if (c==null) if (c==null)
...@@ -1270,6 +1432,7 @@ public class Arrays { ...@@ -1270,6 +1432,7 @@ public class Arrays {
* low is the index in dest to start sorting * low is the index in dest to start sorting
* high is the end index in dest to end sorting * high is the end index in dest to end sorting
* off is the offset into src corresponding to low in dest * off is the offset into src corresponding to low in dest
* To be removed in a future release.
*/ */
private static void mergeSort(Object[] src, private static void mergeSort(Object[] src,
Object[] dest, Object[] dest,
......
src/share/classes/java/util/Collections.java
浏览文件 @ fb246063
...@@ -100,23 +100,42 @@ public class Collections { ...@@ -100,23 +100,42 @@ public class Collections {
/** /**
* Sorts the specified list into ascending order, according to the * Sorts the specified list into ascending order, according to the
* <i>natural ordering</i> of its elements. All elements in the list must * {@linkplain Comparable natural ordering} of its elements.
* implement the <tt>Comparable</tt> interface. Furthermore, all elements * All elements in the list must implement the {@link Comparable}
* in the list must be <i>mutually comparable</i> (that is, * interface. Furthermore, all elements in the list must be
* <tt>e1.compareTo(e2)</tt> must not throw a <tt>ClassCastException</tt> * <i>mutually comparable</i> (that is, {@code e1.compareTo(e2)}
* for any elements <tt>e1</tt> and <tt>e2</tt> in the list).<p> * must not throw a {@code ClassCastException} for any elements
* * {@code e1} and {@code e2} in the list).
* This sort is guaranteed to be <i>stable</i>: equal elements will *
* not be reordered as a result of the sort.<p> * <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* * not be reordered as a result of the sort.
* The specified list must be modifiable, but need not be resizable.<p> *
* * <p>The specified list must be modifiable, but need not be resizable.
* The sorting algorithm is a modified mergesort (in which the merge is *
* omitted if the highest element in the low sublist is less than the * <p>Implementation note: This implementation is a stable, adaptive,
* lowest element in the high sublist). This algorithm offers guaranteed * iterative mergesort that requires far fewer than n lg(n) comparisons
* n log(n) performance. * when the input array is partially sorted, while offering the
* * performance of a traditional mergesort when the input array is
* This implementation dumps the specified list into an array, sorts * randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
*
* <p>The implementation takes equal advantage of ascending and
* descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
*
* <p>The implementation was adapted from Tim Peters's list sort for Python
* (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
*
* <p>This implementation dumps the specified list into an array, sorts
* the array, and iterates over the list resetting each element * the array, and iterates over the list resetting each element
* from the corresponding position in the array. This avoids the * from the corresponding position in the array. This avoids the
* n<sup>2</sup> log(n) performance that would result from attempting * n<sup>2</sup> log(n) performance that would result from attempting
...@@ -126,8 +145,10 @@ public class Collections { ...@@ -126,8 +145,10 @@ public class Collections {
* @throws ClassCastException if the list contains elements that are not * @throws ClassCastException if the list contains elements that are not
* <i>mutually comparable</i> (for example, strings and integers). * <i>mutually comparable</i> (for example, strings and integers).
* @throws UnsupportedOperationException if the specified list's * @throws UnsupportedOperationException if the specified list's
* list-iterator does not support the <tt>set</tt> operation. * list-iterator does not support the {@code set} operation.
* @see Comparable * @throws IllegalArgumentException (optional) if the implementation
* detects that the natural ordering of the list elements is
* found to violate the {@link Comparable} contract
*/ */
public static <T extends Comparable<? super T>> void sort(List<T> list) { public static <T extends Comparable<? super T>> void sort(List<T> list) {
Object[] a = list.toArray(); Object[] a = list.toArray();
...@@ -143,19 +164,38 @@ public class Collections { ...@@ -143,19 +164,38 @@ public class Collections {
* Sorts the specified list according to the order induced by the * Sorts the specified list according to the order induced by the
* specified comparator. All elements in the list must be <i>mutually * specified comparator. All elements in the list must be <i>mutually
* comparable</i> using the specified comparator (that is, * comparable</i> using the specified comparator (that is,
* <tt>c.compare(e1, e2)</tt> must not throw a <tt>ClassCastException</tt> * {@code c.compare(e1, e2)} must not throw a {@code ClassCastException}
* for any elements <tt>e1</tt> and <tt>e2</tt> in the list).<p> * for any elements {@code e1} and {@code e2} in the list).
* *
* This sort is guaranteed to be <i>stable</i>: equal elements will * <p>This sort is guaranteed to be <i>stable</i>: equal elements will
* not be reordered as a result of the sort.<p> * not be reordered as a result of the sort.
* *
* The sorting algorithm is a modified mergesort (in which the merge is * <p>The specified list must be modifiable, but need not be resizable.
* omitted if the highest element in the low sublist is less than the *
* lowest element in the high sublist). This algorithm offers guaranteed * <p>Implementation note: This implementation is a stable, adaptive,
* n log(n) performance. * iterative mergesort that requires far fewer than n lg(n) comparisons
* * when the input array is partially sorted, while offering the
* The specified list must be modifiable, but need not be resizable. * performance of a traditional mergesort when the input array is
* This implementation dumps the specified list into an array, sorts * randomly ordered. If the input array is nearly sorted, the
* implementation requires approximately n comparisons. Temporary
* storage requirements vary from a small constant for nearly sorted
* input arrays to n/2 object references for randomly ordered input
* arrays.
*
* <p>The implementation takes equal advantage of ascending and
* descending order in its input array, and can take advantage of
* ascending and descending order in different parts of the the same
* input array. It is well-suited to merging two or more sorted arrays:
* simply concatenate the arrays and sort the resulting array.
*
* <p>The implementation was adapted from Tim Peters's list sort for Python
* (<a href="http://svn.python.org/projects/python/trunk/Objects/listsort.txt">
* TimSort</a>). It uses techiques from Peter McIlroy's "Optimistic
* Sorting and Information Theoretic Complexity", in Proceedings of the
* Fourth Annual ACM-SIAM Symposium on Discrete Algorithms, pp 467-474,
* January 1993.
*
* <p>This implementation dumps the specified list into an array, sorts
* the array, and iterates over the list resetting each element * the array, and iterates over the list resetting each element
* from the corresponding position in the array. This avoids the * from the corresponding position in the array. This avoids the
* n<sup>2</sup> log(n) performance that would result from attempting * n<sup>2</sup> log(n) performance that would result from attempting
...@@ -163,13 +203,14 @@ public class Collections { ...@@ -163,13 +203,14 @@ public class Collections {
* *
* @param list the list to be sorted. * @param list the list to be sorted.
* @param c the comparator to determine the order of the list. A * @param c the comparator to determine the order of the list. A
* <tt>null</tt> value indicates that the elements' <i>natural * {@code null} value indicates that the elements' <i>natural
* ordering</i> should be used. * ordering</i> should be used.
* @throws ClassCastException if the list contains elements that are not * @throws ClassCastException if the list contains elements that are not
* <i>mutually comparable</i> using the specified comparator. * <i>mutually comparable</i> using the specified comparator.
* @throws UnsupportedOperationException if the specified list's * @throws UnsupportedOperationException if the specified list's
* list-iterator does not support the <tt>set</tt> operation. * list-iterator does not support the {@code set} operation.
* @see Comparator * @throws IllegalArgumentException (optional) if the comparator is
* found to violate the {@link Comparator} contract
*/ */
public static <T> void sort(List<T> list, Comparator<? super T> c) { public static <T> void sort(List<T> list, Comparator<? super T> c) {
Object[] a = list.toArray(); Object[] a = list.toArray();
......
src/share/classes/java/util/ComparableTimSort.java 0 → 100644
浏览文件 @ fb246063
/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Sun designates this
* particular file as subject to the "Classpath" exception as provided
* by Sun in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
* CA 95054 USA or visit www.sun.com if you need additional information or
* have any questions.
*/
package java.util;
/**
* This is a near duplicate of {@link TimSort}, modified for use with
* arrays of objects that implement {@link Comparable}, instead of using
* explicit comparators.
*
* <p>If you are using an optimizing VM, you may find that ComparableTimSort
* offers no performance benefit over TimSort in conjunction with a
* comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
* If this is the case, you are better off deleting ComparableTimSort to
* eliminate the code duplication. (See Arrays.java for details.)
*
* @author Josh Bloch
*/
class ComparableTimSort {
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* {@link #minRunLength} computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private static final int MIN_MERGE = 32;
/**
* The array being sorted.
*/
private final Object[] a;
/**
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
*/
private static final int MIN_GALLOP = 7;
/**
* This controls when we get *into* galloping mode. It is initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* random data, and lower for highly structured data.
*/
private int minGallop = MIN_GALLOP;
/**
* Maximum initial size of tmp array, which is used for merging. The array
* can grow to accommodate demand.
*
* Unlike Tim's original C version, we do not allocate this much storage
* when sorting smaller arrays. This change was required for performance.
*/
private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
/**
* Temp storage for merges.
*/
private Object[] tmp;
/**
* A stack of pending runs yet to be merged. Run i starts at
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that:
*
* runBase[i] + runLen[i] == runBase[i + 1]
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
private int stackSize = 0; // Number of pending runs on stack
private final int[] runBase;
private final int[] runLen;
/**
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
* @param a the array to be sorted
*/
private ComparableTimSort(Object[] a) {
this.a = a;
// Allocate temp storage (which may be increased later if necessary)
int len = a.length;
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
Object[] newArray = new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
tmp = newArray;
/*
* Allocate runs-to-be-merged stack (which cannot be expanded). The
* stack length requirements are described in listsort.txt. The C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays (e.g.,
* 100 elements) in Java. Therefore, we use smaller (but sufficiently
* large) stack lengths for smaller arrays. The "magic numbers" in the
* computation below must be changed if MIN_MERGE is decreased. See
* the MIN_MERGE declaration above for more information.
*/
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 119151 ? 19 : 40);
runBase = new int[stackLen];
runLen = new int[stackLen];
}
/*
* The next two methods (which are package private and static) constitute
* the entire API of this class. Each of these methods obeys the contract
* of the public method with the same signature in java.util.Arrays.
*/
static void sort(Object[] a) {
sort(a, 0, a.length);
}
static void sort(Object[] a, int lo, int hi) {
rangeCheck(a.length, lo, hi);
int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi);
binarySort(a, lo, hi, lo + initRunLen);
return;
}
/**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
*/
ComparableTimSort ts = new ComparableTimSort(a);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi);
// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen);
runLen = force;
}
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
assert lo == hi;
ts.mergeForceCollapse();
assert ts.stackSize == 1;
}
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be sorted
* @param hi the index after the last element in the range to be sorted
* @param start the index of the first element in the range that is
* not already known to be sorted (@code lo <= start <= hi}
*/
@SuppressWarnings("fallthrough")
private static void binarySort(Object[] a, int lo, int hi, int start) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
@SuppressWarnings("unchecked")
Comparable<Object> pivot = (Comparable) a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (pivot.compareTo(a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
/*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
switch(n) {
case 2: a[left + 2] = a[left + 1];
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n);
}
a[left] = pivot;
}
}
/**
* Returns the length of the run beginning at the specified position in
* the specified array and reverses the run if it is descending (ensuring
* that the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
*
* @param a the array in which a run is to be counted and possibly reversed
* @param lo index of the first element in the run
* @param hi index after the last element that may be contained in the run.
It is required that @code{lo < hi}.
* @return the length of the run beginning at the specified position in
* the specified array
*/
@SuppressWarnings("unchecked")
private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
while(runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
runHi++;
}
return runHi - lo;
}
/**
* Reverse the specified range of the specified array.
*
* @param a the array in which a range is to be reversed
* @param lo the index of the first element in the range to be reversed
* @param hi the index after the last element in the range to be reversed
*/
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
}
/**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
* is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
/**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
}
/**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
} else {
break; // Invariant is established
}
}
}
/**
* Merges all runs on the stack until only one remains. This method is
* called once, to complete the sort.
*/
private void mergeForceCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
}
}
/**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
@SuppressWarnings("unchecked")
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3;
int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2;
/*
* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
base2, len2, len2 - 1);
assert len2 >= 0;
if (len2 == 0)
return;
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
}
/**
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
* pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
* In other words, key belongs at index b + k; or in other words,
* the first k elements of a should precede key, and the last n - k
* should follow it.
*/
private static int gallopLeft(Comparable<Object> key, Object[] a,
int base, int len, int hint) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (key.compareTo(a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to base
lastOfs += hint;
ofs += hint;
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
final int maxOfs = hint + 1;
while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to base
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
* to the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
if (key.compareTo(a[base + m]) > 0)
lastOfs = m + 1; // a[base + m] < key
else
ofs = m; // key <= a[base + m]
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
}
/**
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private static int gallopRight(Comparable<Object> key, Object[] a,
int base, int len, int hint) {
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (key.compareTo(a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
lastOfs += hint;
ofs += hint;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
if (key.compareTo(a[base + m]) < 0)
ofs = m; // key < a[b + m]
else
lastOfs = m + 1; // a[b + m] <= key
}
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
}
/**
* Merges two adjacent runs in place, in a stable fashion. The first
* element of the first run must be greater than the first element of the
* second run (a[base1] > a[base2]), and the last element of the first run
* (a[base1 + len1-1]) must be greater than all elements of the second run.
*
* For performance, this method should be called only when len1 <= len2;
* its twin, mergeHi should be called if len1 >= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
@SuppressWarnings("unchecked")
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
Object[] a = this.a; // For performance
Object[] tmp = ensureCapacity(len1);
System.arraycopy(a, base1, tmp, 0, len1);
int cursor1 = 0; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
// Move first element of second run and deal with degenerate cases
a[dest++] = a[cursor2++];
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
return;
}
int minGallop = this.minGallop; // Use local variable for performance
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
assert len1 > 1 && len2 > 0;
if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 1 && len2 > 0;
count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
a[dest++] = a[cursor2++];
if (--len2 == 0)
break outer;
count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0);
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0)
break outer;
}
a[dest++] = tmp[cursor1++];
if (--len1 == 1)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len1 == 1) {
assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len2 == 0;
assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
}
/**
* Like mergeLo, except that this method should be called only if
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
@SuppressWarnings("unchecked")
private void mergeHi(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
Object[] a = this.a; // For performance
Object[] tmp = ensureCapacity(len2);
System.arraycopy(a, base2, tmp, 0, len2);
int cursor1 = base1 + len1 - 1; // Indexes into a
int cursor2 = len2 - 1; // Indexes into tmp array
int dest = base2 + len2 - 1; // Indexes into a
// Move last element of first run and deal with degenerate cases
a[dest--] = a[cursor1--];
if (--len1 == 0) {
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2];
return;
}
int minGallop = this.minGallop; // Use local variable for performance
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
assert len1 > 0 && len2 > 1;
if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) {
a[dest--] = a[cursor1--];
count1++;
count2 = 0;
if (--len1 == 0)
break outer;
} else {
a[dest--] = tmp[cursor2--];
count2++;
count1 = 0;
if (--len2 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 0 && len2 > 1;
count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
if (len1 == 0)
break outer;
}
a[dest--] = tmp[cursor2--];
if (--len2 == 1)
break outer;
count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, 0, len2, len2 - 1);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1)
break outer; // len2 == 1 || len2 == 0
}
a[dest--] = a[cursor1--];
if (--len1 == 0)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len2 == 1) {
assert len1 > 0;
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
} else if (len2 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len1 == 0;
assert len2 > 0;
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
}
}
/**
* Ensures that the external array tmp has at least the specified
* number of elements, increasing its size if necessary. The size
* increases exponentially to ensure amortized linear time complexity.
*
* @param minCapacity the minimum required capacity of the tmp array
* @return tmp, whether or not it grew
*/
private Object[] ensureCapacity(int minCapacity) {
if (tmp.length < minCapacity) {
// Compute smallest power of 2 > minCapacity
int newSize = minCapacity;
newSize |= newSize >> 1;
newSize |= newSize >> 2;
newSize |= newSize >> 4;
newSize |= newSize >> 8;
newSize |= newSize >> 16;
newSize++;
if (newSize < 0) // Not bloody likely!
newSize = minCapacity;
else
newSize = Math.min(newSize, a.length >>> 1);
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
Object[] newArray = new Object[newSize];
tmp = newArray;
}
return tmp;
}
/**
* Checks that fromIndex and toIndex are in range, and throws an
* appropriate exception if they aren't.
*
* @param arrayLen the length of the array
* @param fromIndex the index of the first element of the range
* @param toIndex the index after the last element of the range
* @throws IllegalArgumentException if fromIndex > toIndex
* @throws ArrayIndexOutOfBoundsException if fromIndex < 0
* or toIndex > arrayLen
*/
private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
if (fromIndex > toIndex)
throw new IllegalArgumentException("fromIndex(" + fromIndex +
") > toIndex(" + toIndex+")");
if (fromIndex < 0)
throw new ArrayIndexOutOfBoundsException(fromIndex);
if (toIndex > arrayLen)
throw new ArrayIndexOutOfBoundsException(toIndex);
}
}
src/share/classes/java/util/TimSort.java 0 → 100644
浏览文件 @ fb246063
/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Sun designates this
* particular file as subject to the "Classpath" exception as provided
* by Sun in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
* CA 95054 USA or visit www.sun.com if you need additional information or
* have any questions.
*/
package java.util;
/**
* A stable, adaptive, iterative mergesort that requires far fewer than
* n lg(n) comparisons when running on partially sorted arrays, while
* offering performance comparable to a traditional mergesort when run
* on random arrays. Like all proper mergesorts, this sort is stable and
* runs O(n log n) time (worst case). In the worst case, this sort requires
* temporary storage space for n/2 object references; in the best case,
* it requires only a small constant amount of space.
*
* This implementation was adapted from Tim Peters's list sort for
* Python, which is described in detail here:
*
* http://svn.python.org/projects/python/trunk/Objects/listsort.txt
*
* Tim's C code may be found here:
*
* http://svn.python.org/projects/python/trunk/Objects/listobject.c
*
* The underlying techniques are described in this paper (and may have
* even earlier origins):
*
* "Optimistic Sorting and Information Theoretic Complexity"
* Peter McIlroy
* SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
* pp 467-474, Austin, Texas, 25-27 January 1993.
*
* While the API to this class consists solely of static methods, it is
* (privately) instantiable; a TimSort instance holds the state of an ongoing
* sort, assuming the input array is large enough to warrant the full-blown
* TimSort. Small arrays are sorted in place, using a binary insertion sort.
*
* @author Josh Bloch
*/
class TimSort<T> {
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* {@link #minRunLength} computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private static final int MIN_MERGE = 32;
/**
* The array being sorted.
*/
private final T[] a;
/**
* The comparator for this sort.
*/
private final Comparator<? super T> c;
/**
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
*/
private static final int MIN_GALLOP = 7;
/**
* This controls when we get *into* galloping mode. It is initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* random data, and lower for highly structured data.
*/
private int minGallop = MIN_GALLOP;
/**
* Maximum initial size of tmp array, which is used for merging. The array
* can grow to accommodate demand.
*
* Unlike Tim's original C version, we do not allocate this much storage
* when sorting smaller arrays. This change was required for performance.
*/
private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
/**
* Temp storage for merges.
*/
private T[] tmp; // Actual runtime type will be Object[], regardless of T
/**
* A stack of pending runs yet to be merged. Run i starts at
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that:
*
* runBase[i] + runLen[i] == runBase[i + 1]
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
private int stackSize = 0; // Number of pending runs on stack
private final int[] runBase;
private final int[] runLen;
/**
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
* @param a the array to be sorted
* @param c the comparator to determine the order of the sort
*/
private TimSort(T[] a, Comparator<? super T> c) {
this.a = a;
this.c = c;
// Allocate temp storage (which may be increased later if necessary)
int len = a.length;
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
tmp = newArray;
/*
* Allocate runs-to-be-merged stack (which cannot be expanded). The
* stack length requirements are described in listsort.txt. The C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays (e.g.,
* 100 elements) in Java. Therefore, we use smaller (but sufficiently
* large) stack lengths for smaller arrays. The "magic numbers" in the
* computation below must be changed if MIN_MERGE is decreased. See
* the MIN_MERGE declaration above for more information.
*/
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 119151 ? 19 : 40);
runBase = new int[stackLen];
runLen = new int[stackLen];
}
/*
* The next two methods (which are package private and static) constitute
* the entire API of this class. Each of these methods obeys the contract
* of the public method with the same signature in java.util.Arrays.
*/
static <T> void sort(T[] a, Comparator<? super T> c) {
sort(a, 0, a.length, c);
}
static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {
if (c == null) {
Arrays.sort(a, lo, hi);
return;
}
rangeCheck(a.length, lo, hi);
int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
return;
}
/**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
*/
TimSort<T> ts = new TimSort<T>(a, c);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen, c);
runLen = force;
}
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
assert lo == hi;
ts.mergeForceCollapse();
assert ts.stackSize == 1;
}
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be sorted
* @param hi the index after the last element in the range to be sorted
* @param start the index of the first element in the range that is
* not already known to be sorted (@code lo <= start <= hi}
* @param c comparator to used for the sort
*/
@SuppressWarnings("fallthrough")
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
T pivot = a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (c.compare(pivot, a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
/*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
switch(n) {
case 2: a[left + 2] = a[left + 1];
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n);
}
a[left] = pivot;
}
}
/**
* Returns the length of the run beginning at the specified position in
* the specified array and reverses the run if it is descending (ensuring
* that the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
*
* @param a the array in which a run is to be counted and possibly reversed
* @param lo index of the first element in the run
* @param hi index after the last element that may be contained in the run.
It is required that @code{lo < hi}.
* @param c the comparator to used for the sort
* @return the length of the run beginning at the specified position in
* the specified array
*/
private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator<? super T> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
while(runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}
return runHi - lo;
}
/**
* Reverse the specified range of the specified array.
*
* @param a the array in which a range is to be reversed
* @param lo the index of the first element in the range to be reversed
* @param hi the index after the last element in the range to be reversed
*/
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
}
/**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
* is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
/**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
}
/**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
} else {
break; // Invariant is established
}
}
}
/**
* Merges all runs on the stack until only one remains. This method is
* called once, to complete the sort.
*/
private void mergeForceCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
}
}
/**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3;
int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2;
/*
* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
int k = gallopRight(a[base2], a, base1, len1, 0, c);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
if (len2 == 0)
return;
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
}
/**
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
* pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
* In other words, key belongs at index b + k; or in other words,
* the first k elements of a should precede key, and the last n - k
* should follow it.
*/
private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (c.compare(key, a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to base
lastOfs += hint;
ofs += hint;
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
final int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to base
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
* to the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
if (c.compare(key, a[base + m]) > 0)
lastOfs = m + 1; // a[base + m] < key
else
ofs = m; // key <= a[base + m]
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
}
/**
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* The closer hint is to the result, the faster this method will run.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private static <T> int gallopRight(T key, T[] a, int base, int len,
int hint, Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (c.compare(key, a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
lastOfs += hint;
ofs += hint;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
if (c.compare(key, a[base + m]) < 0)
ofs = m; // key < a[b + m]
else
lastOfs = m + 1; // a[b + m] <= key
}
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
}
/**
* Merges two adjacent runs in place, in a stable fashion. The first
* element of the first run must be greater than the first element of the
* second run (a[base1] > a[base2]), and the last element of the first run
* (a[base1 + len1-1]) must be greater than all elements of the second run.
*
* For performance, this method should be called only when len1 <= len2;
* its twin, mergeHi should be called if len1 >= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len1);
System.arraycopy(a, base1, tmp, 0, len1);
int cursor1 = 0; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
// Move first element of second run and deal with degenerate cases
a[dest++] = a[cursor2++];
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
return;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
assert len1 > 1 && len2 > 0;
if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 1 && len2 > 0;
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
a[dest++] = a[cursor2++];
if (--len2 == 0)
break outer;
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0)
break outer;
}
a[dest++] = tmp[cursor1++];
if (--len1 == 1)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len1 == 1) {
assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len2 == 0;
assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
}
/**
* Like mergeLo, except that this method should be called only if
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
private void mergeHi(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len2);
System.arraycopy(a, base2, tmp, 0, len2);
int cursor1 = base1 + len1 - 1; // Indexes into a
int cursor2 = len2 - 1; // Indexes into tmp array
int dest = base2 + len2 - 1; // Indexes into a
// Move last element of first run and deal with degenerate cases
a[dest--] = a[cursor1--];
if (--len1 == 0) {
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2];
return;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
assert len1 > 0 && len2 > 1;
if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
a[dest--] = a[cursor1--];
count1++;
count2 = 0;
if (--len1 == 0)
break outer;
} else {
a[dest--] = tmp[cursor2--];
count2++;
count1 = 0;
if (--len2 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 0 && len2 > 1;
count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
if (len1 == 0)
break outer;
}
a[dest--] = tmp[cursor2--];
if (--len2 == 1)
break outer;
count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) // len2 == 1 || len2 == 0
break outer;
}
a[dest--] = a[cursor1--];
if (--len1 == 0)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len2 == 1) {
assert len1 > 0;
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
} else if (len2 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len1 == 0;
assert len2 > 0;
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
}
}
/**
* Ensures that the external array tmp has at least the specified
* number of elements, increasing its size if necessary. The size
* increases exponentially to ensure amortized linear time complexity.
*
* @param minCapacity the minimum required capacity of the tmp array
* @return tmp, whether or not it grew
*/
private T[] ensureCapacity(int minCapacity) {
if (tmp.length < minCapacity) {
// Compute smallest power of 2 > minCapacity
int newSize = minCapacity;
newSize |= newSize >> 1;
newSize |= newSize >> 2;
newSize |= newSize >> 4;
newSize |= newSize >> 8;
newSize |= newSize >> 16;
newSize++;
if (newSize < 0) // Not bloody likely!
newSize = minCapacity;
else
newSize = Math.min(newSize, a.length >>> 1);
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[]) new Object[newSize];
tmp = newArray;
}
return tmp;
}
/**
* Checks that fromIndex and toIndex are in range, and throws an
* appropriate exception if they aren't.
*
* @param arrayLen the length of the array
* @param fromIndex the index of the first element of the range
* @param toIndex the index after the last element of the range
* @throws IllegalArgumentException if fromIndex > toIndex
* @throws ArrayIndexOutOfBoundsException if fromIndex < 0
* or toIndex > arrayLen
*/
private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
if (fromIndex > toIndex)
throw new IllegalArgumentException("fromIndex(" + fromIndex +
") > toIndex(" + toIndex+")");
if (fromIndex < 0)
throw new ArrayIndexOutOfBoundsException(fromIndex);
if (toIndex > arrayLen)
throw new ArrayIndexOutOfBoundsException(toIndex);
}
}
test/java/util/TimSort/ArrayBuilder.java 0 → 100644
浏览文件 @ fb246063
/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
* CA 95054 USA or visit www.sun.com if you need additional information or
* have any questions.
*/
import java.util.Random;
import java.math.BigInteger;
public enum ArrayBuilder {
// These seven are from Tim's paper (listsort.txt)
RANDOM_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = rnd.nextInt();
return result;
}
},
DESCENDING_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = len - i;
return result;
}
},
ASCENDING_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = i;
return result;
}
},
ASCENDING_3_RND_EXCH_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = i;
for (int i = 0; i < 3; i++)
swap(result, rnd.nextInt(result.length),
rnd.nextInt(result.length));
return result;
}
},
ASCENDING_10_RND_AT_END_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
int endStart = len - 10;
for (int i = 0; i < endStart; i++)
result[i] = i;
for (int i = endStart; i < len; i++)
result[i] = rnd.nextInt(endStart + 10);
return result;
}
},
ALL_EQUAL_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = 666;
return result;
}
},
DUPS_GALORE_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = rnd.nextInt(4);
return result;
}
},
RANDOM_WITH_DUPS_INT {
public Object[] build(int len) {
Integer[] result = new Integer[len];
for (int i = 0; i < len; i++)
result[i] = rnd.nextInt(len);
return result;
}
},
PSEUDO_ASCENDING_STRING {
public String[] build(int len) {
String[] result = new String[len];
for (int i = 0; i < len; i++)
result[i] = Integer.toString(i);
return result;
}
},
RANDOM_BIGINT {
public BigInteger[] build(int len) {
BigInteger[] result = new BigInteger[len];
for (int i = 0; i < len; i++)
result[i] = HUGE.add(BigInteger.valueOf(rnd.nextInt(len)));
return result;
}
};
public abstract Object[] build(int len);
public void resetRandom() {
rnd = new Random(666);
}
private static Random rnd = new Random(666);
private static void swap(Object[] a, int i, int j) {
Object t = a[i];
a[i] = a[j];
a[j] = t;
}
private static BigInteger HUGE = BigInteger.ONE.shiftLeft(100);
}
test/java/util/TimSort/README 0 → 100644
浏览文件 @ fb246063
This directory contains benchmark programs used to compare the
performance of the TimSort algorithm against the historic 1997
implementation of Arrays.sort. Any future benchmarking will require
minor modifications.
test/java/util/TimSort/SortPerf.java 0 → 100644
浏览文件 @ fb246063
/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
* CA 95054 USA or visit www.sun.com if you need additional information or
* have any questions.
*/
import java.util.Arrays;
public class SortPerf {
private static final int NUM_SETS = 5;
private static final int[] lengths = { 10, 100, 1000, 10000, 1000000 };
// Returns the number of repetitions as a function of the list length
private static int reps(int n) {
return (int) (12000000 / (n * Math.log10(n)));
}
public static void main(String[] args) {
Sorter.warmup();
System.out.print("Strategy,Length");
for (Sorter sorter : Sorter.values())
System.out.print("," + sorter);
System.out.println();
for (ArrayBuilder ab : ArrayBuilder.values()) {
for (int n : lengths) {
System.out.printf("%s,%d", ab, n);
int reps = reps(n);
Object[] proto = ab.build(n);
for (Sorter sorter : Sorter.values()) {
double minTime = Double.POSITIVE_INFINITY;
for (int set = 0; set < NUM_SETS; set++) {
long startTime = System.nanoTime();
for (int k = 0; k < reps; k++) {
Object[] a = proto.clone();
sorter.sort(a);
}
long endTime = System.nanoTime();
double time = (endTime - startTime) / (1000000. * reps);
minTime = Math.min(minTime, time);
}
System.out.printf(",%5f", minTime);
}
System.out.println();
}
}
}
}
test/java/util/TimSort/Sorter.java 0 → 100644
浏览文件 @ fb246063
/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
* CA 95054 USA or visit www.sun.com if you need additional information or
* have any questions.
*/
import java.util.*;
public enum Sorter {
TIMSORT {
public void sort(Object[] array) {
ComparableTimSort.sort(array);
}
},
MERGESORT {
public void sort(Object[] array) {
Arrays.sort(array);
}
};
public abstract void sort(Object[] array);
public static void warmup() {
System.out.println("start warm up");
Integer[] gold = new Integer[10000];
Random random = new java.util.Random();
for (int i=0; i < gold.length; i++)
gold[i] = random.nextInt();
for (int i=0; i < 10000; i++) {
for (Sorter s : values()) {
Integer[] test= gold.clone();
s.sort(test);
}
}
System.out.println(" end warm up");
}
}
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