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体验新版 GitCode,发现更多精彩内容 >>
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0765a89f
编写于
3月 03, 2019
作者:
L
liu13
浏览文件
操作
浏览文件
下载
电子邮件补丁
差异文件
20190303
上级
523ae792
变更
8
隐藏空白更改
内联
并排
Showing
8 changed file
with
128 addition
and
4 deletion
+128
-4
code/lc138.java
code/lc138.java
+1
-1
code/lc242.java
code/lc242.java
+2
-2
code/lc328.java
code/lc328.java
+32
-0
code/lc334.java
code/lc334.java
+21
-0
code/lc341.java
code/lc341.java
+51
-0
code/lc347.java
code/lc347.java
+1
-1
code/lc378.java
code/lc378.java
+14
-0
readme.md
readme.md
+6
-0
未找到文件。
code/lc138.java
浏览文件 @
0765a89f
...
...
@@ -31,7 +31,7 @@ public class lc138 {
node
=
head
;
RandomListNode
res
=
head
.
next
;
RandomListNode
temp
=
head
.
next
;
while
(
temp
.
next
!=
null
){
while
(
temp
!=
null
&&
temp
.
next
!=
null
){
node
.
next
=
temp
.
next
;
temp
.
next
=
node
.
next
.
next
;
node
=
node
.
next
;
...
...
code/lc242.java
浏览文件 @
0765a89f
...
...
@@ -4,8 +4,8 @@ package code;
* 题意:字符串t是否为s打乱后的重排列
* 难度:Easy
* 分类:Hash Table, Sort
* 思路:
* Tips:
* 思路:
hash table 计算即可,<0了直接返回false
* Tips:
和重排列那题联系一下 lc46
*/
public
class
lc242
{
public
boolean
isAnagram
(
String
s
,
String
t
)
{
...
...
code/lc328.java
0 → 100644
浏览文件 @
0765a89f
package
code
;
/*
* 328. Odd Even Linked List
* 题意:奇数的node都在偶数node的后面,第一个节点index为0
* 难度:Medium
* 分类:Linked List
* 思路:while的条件很不好想,记住:这种跳两次的迭代,都判断后边那个节点的 next!=null 为终止条件
* Tips:lc318
*/
public
class
lc328
{
public
class
ListNode
{
int
val
;
ListNode
next
;
ListNode
(
int
x
)
{
val
=
x
;
}
}
public
ListNode
oddEvenList
(
ListNode
head
)
{
if
(
head
==
null
||
head
.
next
==
null
)
return
head
;
ListNode
even
=
head
;
ListNode
odd_head
=
head
.
next
;
ListNode
odd
=
head
.
next
;
while
(
odd
!=
null
&&
odd
.
next
!=
null
){
//这个条件很难想通
even
.
next
=
even
.
next
.
next
;
odd
.
next
=
odd
.
next
.
next
;
even
=
even
.
next
;
odd
=
odd
.
next
;
}
even
.
next
=
odd_head
;
return
head
;
}
}
code/lc334.java
0 → 100644
浏览文件 @
0765a89f
package
code
;
/*
* 334. Increasing Triplet Subsequence
* 题意:数组中是否存在递增的3个数
* 难度:Medium
* 分类:Array
* 思路:思路很清奇,自己没想到,时间空间复杂度都是O(1)。其实和lc300 O(nlgn) 解法思路是一样的,只是固定了dp数组长度为两个数。
* Tips:lc300最长递增子序列
*/
public
class
lc334
{
public
boolean
increasingTriplet
(
int
[]
nums
)
{
int
small
=
Integer
.
MAX_VALUE
;
int
big
=
Integer
.
MAX_VALUE
;
for
(
int
i
=
0
;
i
<
nums
.
length
;
i
++)
{
if
(
nums
[
i
]<
small
)
small
=
nums
[
i
];
else
if
(
nums
[
i
]<
big
)
big
=
nums
[
i
];
else
return
true
;
}
return
false
;
}
}
code/lc341.java
0 → 100644
浏览文件 @
0765a89f
package
code
;
import
java.util.Iterator
;
import
java.util.List
;
import
java.util.Stack
;
public
class
lc341
{
public
interface
NestedInteger
{
// @return true if this NestedInteger holds a single integer, rather than a nested list.
public
boolean
isInteger
();
// @return the single integer that this NestedInteger holds, if it holds a single integer
// Return null if this NestedInteger holds a nested list
public
Integer
getInteger
();
// @return the nested list that this NestedInteger holds, if it holds a nested list
// Return null if this NestedInteger holds a single integer
public
List
<
NestedInteger
>
getList
();
}
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i = new NestedIterator(nestedList);
* while (i.hasNext()) v[f()] = i.next();
*/
public
class
NestedIterator
implements
Iterator
<
Integer
>
{
Stack
<
NestedInteger
>
stack
=
new
Stack
<>();
public
NestedIterator
(
List
<
NestedInteger
>
nestedList
)
{
for
(
int
i
=
nestedList
.
size
()
-
1
;
i
>=
0
;
i
--)
{
stack
.
push
(
nestedList
.
get
(
i
));
}
}
@Override
public
Integer
next
()
{
return
stack
.
pop
().
getInteger
();
}
@Override
public
boolean
hasNext
()
{
while
(!
stack
.
isEmpty
())
{
NestedInteger
curr
=
stack
.
peek
();
if
(
curr
.
isInteger
())
{
return
true
;
}
stack
.
pop
();
for
(
int
i
=
curr
.
getList
().
size
()
-
1
;
i
>=
0
;
i
--)
{
stack
.
push
(
curr
.
getList
().
get
(
i
));
}
}
return
false
;
}
}
}
code/lc347.java
浏览文件 @
0765a89f
...
...
@@ -22,7 +22,7 @@ public class lc347 {
for
(
int
i
=
0
;
i
<
nums
.
length
;
i
++)
{
//放入hashmap计数
hm
.
put
(
nums
[
i
],
hm
.
getOrDefault
(
nums
[
i
],
0
)+
1
);
}
for
(
int
i
:
hm
.
keySet
()
){
//key,value反转,放入treemap
for
(
int
i
:
hm
.
keySet
()
){
//key,value反转,放入treemap
TreeMap中默认是按照升序进行排序的
int
freq
=
hm
.
get
(
i
);
if
(
tm
.
containsKey
(
freq
))
tm
.
get
(
freq
).
add
(
i
);
...
...
code/lc378.java
0 → 100644
浏览文件 @
0765a89f
package
code
;
/*
* 378. Kth Smallest Element in a Sorted Matrix
* 题意:在矩阵中搜索第k大的数,横轴和纵轴都是有序的
* 难度:Medium
* 分类:Binary Search, Heap
* 思路:
* Tips:lc23方法很像
*/
public
class
lc378
{
public
int
kthSmallest
(
int
[][]
matrix
,
int
k
)
{
}
}
readme.md
浏览文件 @
0765a89f
...
...
@@ -127,9 +127,12 @@ Language: Java
| 227
[
Java
](
./code/lc227.java
)
| 230
[
Java
](
./code/lc230.java
)
| 234
[
Java
](
./code/lc234.java
)
| 237
[
Java
](
./code/lc237.java
)
| 238
[
Java
](
./code/lc238.java
)
| 239
[
Java
](
./code/lc239.java
)
| 240
[
Java
](
./code/lc240.java
)
| 242
[
Java
](
./code/lc242.java
)
| 268
[
Java
](
./code/lc268.java
)
| 279
[
Java
](
./code/lc279.java
)
| 283
[
Java
](
./code/lc283.java
)
| 287
[
Java
](
./code/lc287.java
)
...
...
@@ -141,6 +144,9 @@ Language: Java
| 312
[
Java
](
./code/lc312.java
)
| 315
[
Java
](
./code/lc315.java
)
| 322
[
Java
](
./code/lc322.java
)
| 324
[
Java
](
./code/lc324.java
)
| 326
[
Java
](
./code/lc326.java
)
| 329
[
Java
](
./code/lc329.java
)
| 337
[
Java
](
./code/lc337.java
)
| 338
[
Java
](
./code/lc338.java
)
| 347
[
Java
](
./code/lc347.java
)
...
...
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