提交 f2bffe95 编写于 作者: M mduigou

6728865: Provide a better heuristics for Collections.disjoint method

Reviewed-by: alanb, dholmes, chegar, forax
上级 ab869bd6
......@@ -3713,45 +3713,91 @@ public class Collections {
}
/**
* Returns <tt>true</tt> if the two specified collections have no
* Returns {@code true} if the two specified collections have no
* elements in common.
*
* <p>Care must be exercised if this method is used on collections that
* do not comply with the general contract for <tt>Collection</tt>.
* do not comply with the general contract for {@code Collection}.
* Implementations may elect to iterate over either collection and test
* for containment in the other collection (or to perform any equivalent
* computation). If either collection uses a nonstandard equality test
* (as does a {@link SortedSet} whose ordering is not <i>compatible with
* equals</i>, or the key set of an {@link IdentityHashMap}), both
* (as does a {@link SortedSet} whose ordering is not <em>compatible with
* equals</em>, or the key set of an {@link IdentityHashMap}), both
* collections must use the same nonstandard equality test, or the
* result of this method is undefined.
*
* <p>Care must also be exercised when using collections that have
* restrictions on the elements that they may contain. Collection
* implementations are allowed to throw exceptions for any operation
* involving elements they deem ineligible. For absolute safety the
* specified collections should contain only elements which are
* eligible elements for both collections.
*
* <p>Note that it is permissible to pass the same collection in both
* parameters, in which case the method will return true if and only if
* the collection is empty.
* parameters, in which case the method will return {@code true} if and
* only if the collection is empty.
*
* @param c1 a collection
* @param c2 a collection
* @throws NullPointerException if either collection is null
* @return {@code true} if the two specified collections have no
* elements in common.
* @throws NullPointerException if either collection is {@code null}.
* @throws NullPointerException if one collection contains a {@code null}
* element and {@code null} is not an eligible element for the other collection.
* (optional)
* @throws ClassCastException if one collection contains an element that is
* of a type which is ineligible for the other collection. (optional)
* @since 1.5
*/
public static boolean disjoint(Collection<?> c1, Collection<?> c2) {
/*
* We're going to iterate through c1 and test for inclusion in c2.
* If c1 is a Set and c2 isn't, swap the collections. Otherwise,
* place the shorter collection in c1. Hopefully this heuristic
* will minimize the cost of the operation.
*/
if ((c1 instanceof Set) && !(c2 instanceof Set) ||
(c1.size() > c2.size())) {
Collection<?> tmp = c1;
c1 = c2;
c2 = tmp;
// The collection to be used for contains(). Preference is given to
// the collection who's contains() has lower O() complexity.
Collection<?> contains = c2;
// The collection to be iterated. If the collections' contains() impl
// are of different O() complexity, the collection with slower
// contains() will be used for iteration. For collections who's
// contains() are of the same complexity then best performance is
// achieved by iterating the smaller collection.
Collection<?> iterate = c1;
// Performance optimization cases. The heuristics:
// 1. Generally iterate over c1.
// 2. If c1 is a Set then iterate over c2.
// 3. If either collection is empty then result is always true.
// 4. Iterate over the smaller Collection.
if (c1 instanceof Set) {
// Use c1 for contains as a Set's contains() is expected to perform
// better than O(N/2)
iterate = c2;
contains = c1;
} else if (!(c2 instanceof Set)) {
// Both are mere Collections. Iterate over smaller collection.
// Example: If c1 contains 3 elements and c2 contains 50 elements and
// assuming contains() requires ceiling(N/2) comparisons then
// checking for all c1 elements in c2 would require 75 comparisons
// (3 * ceiling(50/2)) vs. checking all c2 elements in c1 requiring
// 100 comparisons (50 * ceiling(3/2)).
int c1size = c1.size();
int c2size = c2.size();
if (c1size == 0 || c2size == 0) {
// At least one collection is empty. Nothing will match.
return true;
}
if (c1size > c2size) {
iterate = c2;
contains = c1;
}
}
for (Object e : c1)
if (c2.contains(e))
for (Object e : iterate) {
if (contains.contains(e)) {
// Found a common element. Collections are not disjoint.
return false;
}
}
// No common elements were found.
return true;
}
......
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