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Merge branch 'youngyangyang04:master' into master

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`@@ -35,7 +35,7 @@`
`35`	`35`
`36`	`36`	`四数之和,和[15.三数之和](https://programmercarl.com/0015.三数之和.html)是一个思路,都是使用双指针法, 基本解法就是在[15.三数之和](https://programmercarl.com/0015.三数之和.html) 的基础上再套一层for循环。`
`37`	`37`
`38`		-但是有一些细节需要注意,例如: 不要判断`nums[k] > target` 就返回了,三数之和可以通过 `nums[i] > 0` 就返回了,因为 0 已经是确定的数了,四数之和这道题目 target是任意值。比如:数组是`[-4, -3, -2, -1]`,`target`是`-10`,不能因为`-4 > -10`而跳过。但是我们依旧可以去做剪枝,逻辑变成`nums[i] > target && (nums[i] >=0 \|\| target >= 0)`就可以了。
	`38`	+但是有一些细节需要注意,例如: 不要判断`nums[k] > target` 就返回了,三数之和可以通过 `nums[i] > 0` 就返回了,因为 0 已经是确定的数了,四数之和这道题目 target是任意值。比如:数组是`[-4, -3, -2, -1]`,`target`是`-10`,不能因为`-4 > -10`而跳过。但是我们依旧可以去做剪枝,逻辑变成`nums[k] > target && (nums[k] >=0 \|\| target >= 0)`就可以了。
`39`	`39`
`40`	`40`	`[15.三数之和](https://programmercarl.com/0015.三数之和.html)的双指针解法是一层for循环num[i]为确定值,然后循环内有left和right下标作为双指针,找到nums[i] + nums[left] + nums[right] == 0。`
`41`	`41`

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`@@ -129,6 +129,36 @@ class Solution {`
`129`	`129`	`}`
`130`	`130`	```
`131`	`131`
	`132`	`+`
	`133`	+```java
	`134`	`+class Solution {`
	`135`	`+ public ListNode removeNthFromEnd(ListNode head, int n) {`
	`136`	`+ // 创建一个新的哑节点,指向原链表头`
	`137`	`+ ListNode s = new ListNode(-1, head);`
	`138`	`+ // 递归调用remove方法,从哑节点开始进行删除操作`
	`139`	`+ remove(s, n);`
	`140`	`+ // 返回新链表的头(去掉可能的哑节点)`
	`141`	`+ return s.next;`
	`142`	`+ }`
	`143`	`+`
	`144`	`+ public int remove(ListNode p, int n) {`
	`145`	`+ // 递归结束条件:如果当前节点为空,返回0`
	`146`	`+ if (p == null) {`
	`147`	`+ return 0;`
	`148`	`+ }`
	`149`	`+ // 递归深入到下一个节点`
	`150`	`+ int net = remove(p.next, n);`
	`151`	`+ // 如果当前节点是倒数第n个节点,进行删除操作`
	`152`	`+ if (net == n) {`
	`153`	`+ p.next = p.next.next;`
	`154`	`+ }`
	`155`	`+ // 返回当前节点的总深度`
	`156`	`+ return net + 1;`
	`157`	`+ }`
	`158`	`+}`
	`159`	+```
	`160`	`+`
	`161`	`+`
`132`	`162`	`### Python:`
`133`	`163`
`134`	`164`	```python

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`@@ -172,6 +172,29 @@ class Solution {`
`172`	`172`	`}`
`173`	`173`	```
`174`	`174`
	`175`	+```java
	`176`	`+// 解法二`
	`177`	`+// 对应的另一半一定在栈顶`
	`178`	`+class Solution {`
	`179`	`+ public boolean isValid(String s) {`
	`180`	`+ Stack<Character> stack = new Stack<>();`
	`181`	`+ for(char c : s.toCharArray()){`
	`182`	`+ // 有对应的另一半就直接消消乐`
	`183`	`+ if(c == ')' && !stack.isEmpty() && stack.peek() == '(')`
	`184`	`+ stack.pop();`
	`185`	`+ else if(c == '}' && !stack.isEmpty() && stack.peek() == '{')`
	`186`	`+ stack.pop();`
	`187`	`+ else if(c == ']' && !stack.isEmpty() && stack.peek() == '[')`
	`188`	`+ stack.pop();`
	`189`	`+ else`
	`190`	`+ stack.push(c);// 没有匹配的就放进去`
	`191`	`+ }`
	`192`	`+`
	`193`	`+ return stack.isEmpty();`
	`194`	`+ }`
	`195`	`+}`
	`196`	+```
	`197`	`+`
`175`	`198`	`### Python:`
`176`	`199`
`177`	`200`	```python

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`@@ -267,7 +267,7 @@ class Solution {`
`267`	`267`	`//利用双端队列手动实现单调队列`
`268`	`268`	`/**`
`269`	`269`	`* 用一个单调队列来存储对应的下标,每当窗口滑动的时候,直接取队列的头部指针对应的值放入结果集即可`
`270`		`- * 单调队列类似 (tail -->) 3 --> 2 --> 1 --> 0 (--> head) (右边为头结点,元素存的是下标)`
	`270`	`+ * 单调递减队列类似 (head -->) 3 --> 2 --> 1 --> 0 (--> tail) (左边为头结点,元素存的是下标)`
`271`	`271`	`*/`
`272`	`272`	`class Solution {`
`273`	`273`	`public int[] maxSlidingWindow(int[] nums, int k) {`
`@@ -281,7 +281,7 @@ class Solution {`
`281`	`281`	`while(!deque.isEmpty() && deque.peek() < i - k + 1){`
`282`	`282`	`deque.poll();`
`283`	`283`	`}`
`284`		`- // 2.既然是单调,就要保证每次放进去的数字要比末尾的都大,否则也弹出`
	`284`	`+ // 2.维护单调递减队列:新元素若大于队尾元素,则弹出队尾元素,直到满足单调性`
`285`	`285`	`while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {`
`286`	`286`	`deque.pollLast();`
`287`	`287`	`}`
`@@ -926,3 +926,4 @@ int* maxSlidingWindow(int* nums, int numsSize, int k, int* returnSize) {`
`926`	`926`	`<a href="https://programmercarl.com/other/kstar.html" target="_blank">`
`927`	`927`	`<img src="../pics/网站星球宣传海报.jpg" width="1000"/>`
`928`	`928`	`</a>`
	`929`	`+`

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