Merge branch 'master' of https://github.com/youngyangyang04/leetcode

problems/0063.不同路径II.md

@@ -232,6 +232,38 @@ class Solution:
         return dp[-1][-1]
 ```
 
+```python
+class Solution:
+    """
+    使用一维dp数组
+    """
+
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        m, n = len(obstacleGrid), len(obstacleGrid[0])
+
+        # 初始化dp数组
+        # 该数组缓存当前行
+        curr = [0] * n
+        for j in range(n):
+            if obstacleGrid[0][j] == 1:
+                break
+            curr[j] = 1
+            
+        for i in range(1, m): # 从第二行开始
+            for j in range(n): # 从第一列开始，因为第一列可能有障碍物
+                # 有障碍物处无法通行，状态就设成0
+                if obstacleGrid[i][j] == 1:
+                    curr[j] = 0
+                elif j > 0:
+                    # 等价于
+                    # dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
+                    curr[j] = curr[j] + curr[j - 1]
+                # 隐含的状态更新
+                # dp[i][0] = dp[i - 1][0]
+        
+        return curr[n - 1]
+```
+
 
 Go：
 

problems/0235.二叉搜索树的最近公共祖先.md

@@ -314,7 +314,62 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
 ```
 
 
+JavaScript版本
+> 递归
 
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+
+/**
+ * @param {TreeNode} root
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {TreeNode}
+ */
+var lowestCommonAncestor = function(root, p, q) {
+    if(root.val > p.val && root.val > q.val)
+        return lowestCommonAncestor(root.left, p , q);
+    else if(root.val < p.val && root.val < q.val) 
+        return lowestCommonAncestor(root.right, p , q);
+    return root;
+};
+```
+
+> 迭代
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+
+/**
+ * @param {TreeNode} root
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {TreeNode}
+ */
+var lowestCommonAncestor = function(root, p, q) {
+    while(1) {
+        if(root.val > p.val && root.val > q.val)
+            root = root.left;
+        else if(root.val < p.val && root.val < q.val)
+            root = root.right;
+        else
+            break;
+    }
+    return root;
+};
+```
 
 
 -----------------------

problems/0239.滑动窗口最大值.md

@@ -208,6 +208,7 @@ public:
 
 Java：
 ```Java
+//解法一
 //自定义数组
 class MyQueue {
     Deque<Integer> deque = new LinkedList<>();
@@ -260,6 +261,40 @@ class Solution {
         return res;
     }
 }
+
+//解法二
+//利用双端队列手动实现单调队列
+/**
+ * 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
+ * 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)
+ */
+class Solution {
+    public int[] maxSlidingWindow(int[] nums, int k) {
+        ArrayDeque<Integer> deque = new ArrayDeque<>();
+        int n = nums.length;
+        int[] res = new int[n - k + 1];
+        int idx = 0;
+        for(int i = 0; i < n; i++) {
+            // 根据题意，i为nums下标，是要在[i - k + 1, k] 中选到最大值，只需要保证两点
+            // 1.队列头结点需要在[i - k + 1, k]范围内，不符合则要弹出
+            while(!deque.isEmpty() && deque.peek() < i - k + 1){
+                deque.poll();
+            }
+            // 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
+            while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
+                deque.pollLast();
+            }
+
+            deque.offer(i);
+
+            // 因为单调，当i增长到符合第一个k范围的时候，每滑动一步都将队列头节点放入结果就行了
+            if(i >= k - 1){
+                res[idx++] = nums[deque.peek()];
+            }
+        }
+        return res;
+    }
+}
 ```
 
 Python：

problems/0344.反转字符串.md

@@ -155,7 +155,7 @@ class Solution {
 ```
 
 Python：
-```python3
+```python
 class Solution:
     def reverseString(self, s: List[str]) -> None:
         """
@@ -166,6 +166,17 @@ class Solution:
             s[left], s[right] = s[right], s[left]
             left += 1
             right -= 1
+            
+# 下面的写法更加简洁，但是都是同样的算法
+# class Solution:
+#     def reverseString(self, s: List[str]) -> None:
+#         """
+#         Do not return anything, modify s in-place instead.
+#         """
+          # 不需要判别是偶数个还是奇数个序列，因为奇数个的时候，中间那个不需要交换就可
+#         for i in range(len(s)//2):
+#             s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i]
+#         return s
 ```
 
 Go：

problems/0406.根据身高重建队列.md

@@ -214,12 +214,9 @@ Python：
 ```python
 class Solution:
     def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
-        people.sort(key=lambda x: (x[0], -x[1]), reverse=True)
+        people.sort(key=lambda x: (-x[0], x[1]))
         que = []
         for p in people:
-            if p[1] > len(que):
-                que.append(p)
-            else:
             que.insert(p[1], p)
         return que
 ```

problems/0450.删除二叉搜索树中的节点.md

@@ -358,6 +358,51 @@ func deleteNode1(root *TreeNode)*TreeNode{
 }
 ```
 
+JavaScript版本
+
+> 递归
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} key
+ * @return {TreeNode}
+ */
+var deleteNode = function (root, key) {
+    if (root === null)
+        return root;
+    if (root.val === key) {
+        if (!root.left)
+            return root.right;
+        else if (!root.right)
+            return root.left;
+        else {
+            let cur = root.right;
+            while (cur.left) {
+                cur = cur.left;
+            }
+            cur.left = root.left;
+            let temp = root;
+            root = root.right;
+            delete root;
+            return root;
+        }
+    }
+    if (root.val > key)
+        root.left = deleteNode(root.left, key);
+    if (root.val < key)
+        root.right = deleteNode(root.right, key);
+    return root;
+};
+```
 
 
 

problems/0617.合并二叉树.md

@@ -368,6 +368,20 @@ func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
         Right: mergeTrees(t1.Right,t2.Right)}
     return root
 }
+
+// 前序遍历简洁版
+func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
+    if root1 == nil {
+        return root2
+    }
+    if root2 == nil {
+        return root1
+    }
+    root1.Val += root2.Val
+    root1.Left = mergeTrees(root1.Left, root2.Left)
+    root1.Right = mergeTrees(root1.Right, root2.Right)
+    return root1
+}
 ```
 
 JavaScript:

problems/0701.二叉搜索树中的插入操作.md

@@ -286,8 +286,118 @@ func insertIntoBST(root *TreeNode, val int) *TreeNode {
 }
 ```
 
+JavaScript版本
 
+> 有返回值的递归写法
 
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} val
+ * @return {TreeNode}
+ */
+var insertIntoBST = function (root, val) {
+    const setInOrder = (root, val) => {
+        if (root === null) {
+            let node = new TreeNode(val);
+            return node;
+        }
+        if (root.val > val)
+            root.left = setInOrder(root.left, val);
+        else if (root.val < val)
+            root.right = setInOrder(root.right, val);
+        return root;
+    }
+    return setInOrder(root, val);
+};
+```
+
+> 无返回值的递归
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} val
+ * @return {TreeNode}
+ */
+var insertIntoBST = function (root, val) {
+    let parent = new TreeNode(0);
+    const preOrder = (cur, val) => {
+        if (cur === null) {
+            let node = new TreeNode(val);
+            if (parent.val > val)
+                parent.left = node;
+            else
+                parent.right = node;
+            return;
+        }
+        parent = cur;
+        if (cur.val > val)
+            preOrder(cur.left, val);
+        if (cur.val < val)
+            preOrder(cur.right, val);
+    }
+    if (root === null)
+        root = new TreeNode(val);
+    preOrder(root, val);
+    return root;
+};
+```
+
+> 迭代
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} val
+ * @return {TreeNode}
+ */
+var insertIntoBST = function (root, val) {
+    if (root === null) {
+        root = new TreeNode(val);
+    } else {
+        let parent = new TreeNode(0);
+        let cur = root;
+        while (cur) {
+            parent = cur;
+            if (cur.val > val)
+                cur = cur.left;
+            else
+                cur = cur.right;
+        }
+        let node = new TreeNode(val);
+        if (parent.val > val)
+            parent.left = node;
+        else
+            parent.right = node;
+    }
+    return root;
+};
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/1047.删除字符串中的所有相邻重复项.md

@@ -127,7 +127,9 @@ Java：
 ```Java
 class Solution {
     public String removeDuplicates(String S) {
-        Deque<Character> deque = new LinkedList<>();
+        //ArrayDeque会比LinkedList在除了删除元素这一点外会快一点
+        //参考：https://stackoverflow.com/questions/6163166/why-is-arraydeque-better-than-linkedlist
+        ArrayDeque<Character> deque = new ArrayDeque<>();
         char ch;
         for (int i = 0; i < S.length(); i++) {
             ch = S.charAt(i);
@@ -171,6 +173,29 @@ class Solution {
 }
 ```
 
+拓展：双指针
+```java
+class Solution {
+    public String removeDuplicates(String s) {
+        char[] ch = s.toCharArray();
+        int fast = 0;
+        int slow = 0;
+        while(fast < s.length()){
+            // 直接用fast指针覆盖slow指针的值
+            ch[slow] = ch[fast];
+            // 遇到前后相同值的，就跳过，即slow指针后退一步，下次循环就可以直接被覆盖掉了
+            if(slow > 0 && ch[slow] == ch[slow - 1]){
+                slow--;
+            }else{
+                slow++;
+            }
+            fast++;
+        }
+        return new String(ch,0,slow);
+    }
+}
+```
+
 Python：
 ```python3
 class Solution:

problems/二叉树的迭代遍历.md

@@ -155,9 +155,82 @@ public:
 
 ## 其他语言版本
 
-
 Java：
 
+```java
+// 前序遍历顺序：中-左-右，入栈顺序：中-右-左
+class Solution {
+    public List<Integer> preorderTraversal(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+        if (root == null){
+            return result;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        stack.push(root);
+        while (!stack.isEmpty()){
+            TreeNode node = stack.pop();
+            result.add(node.val);
+            if (node.right != null){
+                stack.push(node.right);
+            }
+            if (node.left != null){
+                stack.push(node.left);
+            }
+        }
+        return result;
+    }
+}
+
+// 中序遍历顺序: 左-中-右 入栈顺序： 左-右
+class Solution {
+    public List<Integer> inorderTraversal(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+        if (root == null){
+            return result;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode cur = root;
+        while (cur != null || !stack.isEmpty()){
+           if (cur != null){
+               stack.push(cur);
+               cur = cur.left;
+           }else{
+               cur = stack.pop();
+               result.add(cur.val);
+               cur = cur.right;
+           }
+        }
+        return result;
+    }
+}
+
+// 后序遍历顺序 左-右-中 入栈顺序：中-左-右 出栈顺序：中-右-左， 最后翻转结果
+class Solution {
+    public List<Integer> postorderTraversal(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+        if (root == null){
+            return result;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        stack.push(root);
+        while (!stack.isEmpty()){
+            TreeNode node = stack.pop();
+            result.add(node.val);
+            if (node.left != null){
+                stack.push(node.left);
+            }
+            if (node.right != null){
+                stack.push(node.right);
+            }
+        }
+        Collections.reverse(result);
+        return result;
+    }
+}
+```
+
+
+
 
 Python：
 ```python3