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

problems/0005.最长回文子串.md

@@ -288,7 +288,34 @@ class Solution:
         return s[left:right + 1]
 
 ```
+> 双指针法：
+```python
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
 
+        def find_point(i, j, s):
+            while i >= 0 and j < len(s) and s[i] == s[j]:
+                i -= 1
+                j += 1
+            return i + 1, j
+
+        def compare(start, end, left, right):
+            if right - left > end - start:
+                return left, right
+            else:
+                return start, end
+
+        start = 0
+        end = 0
+        for i in range(len(s)):
+            left, right = find_point(i, i, s)
+            start, end = compare(start, end, left, right)
+
+            left, right = find_point(i, i + 1, s)
+            start, end = compare(start, end, left, right)
+        return s[start:end]
+
+```
 ## Go
 
 ```go

problems/0102.二叉树的层序遍历.md

@@ -421,9 +421,10 @@ var levelOrderBottom = function(root) {
             node.left&&queue.push(node.left);
             node.right&&queue.push(node.right);
         }
-        res.push(curLevel);
+	// 从数组前头插入值，避免最后反转数组，减少运算时间
+        res.unshift(curLevel);
     }
-    return res.reverse();
+    return res;
 };
 ```
 
@@ -1263,7 +1264,41 @@ class Solution:
             first = first.left  # 从本层扩展到下一层
         return root
 ```
+JavaScript:
+```javascript
 
+/**
+ * // Definition for a Node.
+ * function Node(val, left, right, next) {
+ *    this.val = val === undefined ? null : val;
+ *    this.left = left === undefined ? null : left;
+ *    this.right = right === undefined ? null : right;
+ *    this.next = next === undefined ? null : next;
+ * };
+ */
+
+/**
+ * @param {Node} root
+ * @return {Node}
+ */
+var connect = function(root) {
+    if (root === null) return root;
+    let queue = [root];
+    while (queue.length) {
+        let n = queue.length;
+        for (let i=0; i<n; i++) {
+            let node = queue.shift();
+            if (i < n-1) {
+                node.next = queue[0];
+            }
+            node.left && queue.push(node.left);
+            node.right && queue.push(node.right);
+        }
+    }
+    return root;
+};
+
+```
 go:
 
 ```GO
@@ -1426,7 +1461,39 @@ class Solution:
             first = dummyHead.next  # 此处为换行操作，更新到下一行
         return root
 ```
+JavaScript:
+```javascript
+/**
+ * // Definition for a Node.
+ * function Node(val, left, right, next) {
+ *    this.val = val === undefined ? null : val;
+ *    this.left = left === undefined ? null : left;
+ *    this.right = right === undefined ? null : right;
+ *    this.next = next === undefined ? null : next;
+ * };
+ */
 
+/**
+ * @param {Node} root
+ * @return {Node}
+ */
+var connect = function(root) {
+    if (root === null) {
+        return null;
+    }
+    let queue = [root];
+    while (queue.length > 0) {
+        let n = queue.length;
+        for (let i=0; i<n; i++) {
+            let node = queue.shift();
+            if (i < n-1) node.next = queue[0];
+            if (node.left != null) queue.push(node.left);
+            if (node.right != null) queue.push(node.right);
+        }
+    }
+    return root;
+};
+```
 go:
 
 ```GO
@@ -1608,6 +1675,36 @@ func maxDepth(root *TreeNode) int {
 
 
 JavaScript：
+```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
+ * @return {number}
+ */
+var maxDepth = function(root) {
+    // 最大的深度就是二叉树的层数
+    if (root === null) return 0;
+    let queue = [root];
+    let height = 0;
+    while (queue.length) {
+        let n = queue.length;
+        height++;
+        for (let i=0; i<n; i++) {
+            let node = queue.shift();
+            node.left && queue.push(node.left);
+            node.right && queue.push(node.right);
+        }
+    }
+    return height;
+};
+```
 
 # 111.二叉树的最小深度
 
@@ -1746,9 +1843,40 @@ func minDepth(root *TreeNode) int {
 }
 ```
 
-
-
 JavaScript：
+```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
+ * @return {number}
+ */
+var minDepth = function(root) {
+    if (root === null) return 0;
+    let queue = [root];
+    let deepth = 0;
+    while (queue.length) {
+        let n = queue.length;
+        deepth++;
+        for (let i=0; i<n; i++) {
+            let node = queue.shift();
+            // 如果左右节点都是null，则该节点深度最小
+            if (node.left === null && node.right === null) {
+                return deepth;
+            }
+            node.left && queue.push(node.left);;
+            node.right && queue.push (node.right);
+        }
+    }
+    return deepth;
+};
+```
 
 
 

problems/0491.递增子序列.md

@@ -313,7 +313,72 @@ var findSubsequences = function(nums) {
 
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+int* length;
+//将当前path中的内容复制到ans中
+void copy() {
+    int* tempPath = (int*)malloc(sizeof(int) * pathTop);
+    memcpy(tempPath, path, pathTop * sizeof(int));
+    length[ansTop] = pathTop;
+    ans[ansTop++] = tempPath;
+}
 
+//查找uset中是否存在值为key的元素
+int find(int* uset, int usetSize, int key) {
+    int i;
+    for(i = 0; i < usetSize; i++) {
+        if(uset[i] == key)
+            return 1;
+    }
+    return 0;
+}
+
+void backTracking(int* nums, int numsSize, int startIndex) {
+    //当path中元素大于1个时，将path拷贝到ans中
+    if(pathTop > 1) {
+        copy();
+    }
+    int* uset = (int*)malloc(sizeof(int) * numsSize);
+    int usetTop = 0;
+    int i;
+    for(i = startIndex; i < numsSize; i++) {
+        //若当前元素小于path中最后一位元素 || 在树的同一层找到了相同的元素，则continue
+        if((pathTop > 0 && nums[i] < path[pathTop - 1]) || find(uset, usetTop, nums[i]))
+            continue;
+        //将当前元素放入uset
+        uset[usetTop++] = nums[i];
+        //将当前元素放入path
+        path[pathTop++] = nums[i];
+        backTracking(nums, numsSize, i + 1);
+        //回溯
+        pathTop--;
+    }
+}
+
+int** findSubsequences(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
+    //辅助数组初始化
+    path = (int*)malloc(sizeof(int) * numsSize);
+    ans = (int**)malloc(sizeof(int*) * 33000);
+    length = (int*)malloc(sizeof(int*) * 33000);
+    pathTop = ansTop = 0;
+
+    backTracking(nums, numsSize, 0);
+
+    //设置数组中返回元素个数，以及每个一维数组的长度
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int i;
+    for(i = 0; i < ansTop; i++) {
+        (*returnColumnSizes)[i] = length[i];
+    }
+    return ans;
+}
+```
 
 
 -----------------------

problems/0617.合并二叉树.md

@@ -507,6 +507,8 @@ func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
 
 ## JavaScript 
 
+> 递归法：
+
 ```javascript
 /**
  * Definition for a binary tree node.
@@ -535,6 +537,53 @@ var mergeTrees = function (root1, root2) {
     return preOrder(root1, root2);
 };
 ```
+> 迭代法：
+
+```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} root1
+ * @param {TreeNode} root2
+ * @return {TreeNode}
+ */
+var mergeTrees = function(root1, root2) {
+    if (root1 === null) return root2;
+    if (root2 === null) return root1;
+
+    let queue = [];
+    queue.push(root1);
+    queue.push(root2);
+    while (queue.length) {
+        let node1 = queue.shift();
+        let node2 = queue.shift();;
+        node1.val += node2.val;
+        if (node1.left !== null && node2.left !== null) {
+            queue.push(node1.left);
+            queue.push(node2.left);
+        }
+        if (node1.right !== null && node2.right !== null) {
+            queue.push(node1.right);
+            queue.push(node2.right);
+        }
+        if (node1.left === null && node2.left !== null) {
+            node1.left = node2.left;
+        }
+        if (node1.right === null && node2.right !== null) {
+            node1.right = node2.right;
+        } 
+    }
+    return root1;
+};
+
+```
 
 
 -----------------------

problems/0704.二分查找.md

@@ -288,15 +288,16 @@ func search(nums []int, target int) int {
  * @param {number} target
  * @return {number}
  */
+/**
 var search = function(nums, target) {
-    let left = 0, right = nums.length;
+    let left = 0, right = nums.length - 1;
-    // 使用左闭右开区间 [left, right)
+    // 使用左闭右闭区间
-    while (left < right) {
+    while (left <= right) {
         let mid = left + Math.floor((right - left)/2);
         if (nums[mid] > target) {
-            right = mid;  // 去左区间寻找
+            right = mid - 1;  // 去左面闭区间寻找
         } else if (nums[mid] < target) {
-            left = mid + 1;   // 去右区间寻找
+            left = mid + 1;   // 去右面闭区间寻找
         } else {
             return mid;
         }

problems/0707.设计链表.md

@@ -880,6 +880,121 @@ MyLinkedList.prototype.deleteAtIndex = function(index) {
  * obj.deleteAtIndex(index)
  */
 ```
+
+TypeScript:
+```TypeScript
+class ListNode {
+    public val: number;
+    public next: ListNode | null;
+    constructor(val?: number, next?: ListNode | null) {
+        this.val = val === undefined ? 0 : val;
+        this.next = next === undefined ? null : next;
+    }
+}
+
+class MyLinkedList {
+    // 记录链表长度
+    private size: number;
+    private head: ListNode | null;
+    private tail: ListNode | null;
+    constructor() {
+        this.size = 0;
+        this.head = null;
+        this.tail = null;
+    }
+
+    // 获取链表中第 index个节点的值
+    get(index: number): number {
+        // 索引无效的情况
+        if (index < 0 || index >= this.size) {
+            return -1;
+        }
+        let curNode = this.getNode(index);
+        // 这里在前置条件下，理论上不会出现 null的情况
+        return curNode.val;
+    }
+
+    // 在链表的第一个元素之前添加一个值为 val的节点。插入后，新节点将成为链表的第一个节点。
+    addAtHead(val: number): void {
+        let node: ListNode = new ListNode(val, this.head);
+        this.head = node;
+        if (!this.tail) {
+            this.tail = node;
+        }
+        this.size++;
+    }
+
+    // 将值为 val 的节点追加到链表的最后一个元素。
+    addAtTail(val: number): void {
+        let node: ListNode = new ListNode(val, null);
+        if (this.tail) {
+            this.tail.next = node;
+        } else {
+            // 还没有尾节点，说明一个节点都还没有
+            this.head = node;
+        }
+        this.tail = node;
+        this.size++;
+    }
+
+    // 在链表中的第 index个节点之前添加值为 val的节点。
+    // 如果 index等于链表的长度，则该节点将附加到链表的末尾。如果 index大于链表长度，则不会插入节点。如果 index小于0，则在头部插入节点。
+    addAtIndex(index: number, val: number): void {
+        if (index === this.size) {
+            this.addAtTail(val);
+            return;
+        }
+        if (index > this.size) {
+            return;
+        }
+        // <= 0 的情况都是在头部插入
+        if (index <= 0) {
+            this.addAtHead(val);
+            return;
+        }
+        // 正常情况
+        // 获取插入位置的前一个 node
+        let curNode = this.getNode(index - 1);
+        let node: ListNode = new ListNode(val, curNode.next);
+        curNode.next = node;
+        this.size++;
+    }
+
+    // 如果索引 index有效，则删除链表中的第 index个节点。
+    deleteAtIndex(index: number): void {
+        if (index < 0 || index >= this.size) {
+            return;
+        }
+        // 处理头节点
+        if (index === 0) {
+            this.head = this.head!.next;
+            this.size--;
+            return;
+        }
+        // 索引有效
+        let curNode: ListNode = this.getNode(index - 1);
+        curNode.next = curNode.next!.next;
+        // 处理尾节点
+        if (index === this.size - 1) {
+            this.tail = curNode;
+        }
+        this.size--;
+    }
+
+    // 获取指定 Node节点
+    private getNode(index: number): ListNode {
+        // 这里不存在没办法获取到节点的情况，都已经在前置方法做过判断
+        // 创建虚拟头节点
+        let curNode: ListNode = new ListNode(0, this.head);
+        for (let i = 0; i <= index; i++) {
+            // 理论上不会出现 null
+            curNode = curNode.next!;
+        }
+        return curNode;
+    }
+}
+```
+
 Kotlin:
 ```kotlin
 class MyLinkedList {

problems/1002.查找常用字符.md

@@ -12,18 +12,24 @@
 
 [力扣题目链接](https://leetcode-cn.com/problems/find-common-characters/)
 
-给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。
+给你一个字符串数组 words ，请你找出所有在 words 的每个字符串中都出现的共用字符（ 包括重复字符），并以数组形式返回。你可以按 任意顺序 返回答案。
 
-你可以按任意顺序返回答案。
+示例 1：
 
-【示例一】
+输入：words = ["bella","label","roller"]
-输入：["bella","label","roller"]
 输出：["e","l","l"]
+示例 2：
 
-【示例二】
+输入：words = ["cool","lock","cook"]
-输入：["cool","lock","cook"]
 输出：["c","o"]
 
+提示：
+
+1 <= words.length <= 100
+1 <= words[i].length <= 100
+words[i] 由小写英文字母组成
+
+
 
 # 思路
 

problems/1035.不相交的线.md

@@ -109,6 +109,41 @@ class Solution:
         return dp[-1][-1]
 ```
 
+
+Golang:
+
+```go
+
+func maxUncrossedLines(A []int, B []int) int {
+	m, n := len(A), len(B)
+	dp := make([][]int, m+1)
+    for i := range dp {
+        dp[i] = make([]int, n+1)
+    }
+
+	for i := 1; i <= len(A); i++ {
+		for j := 1; j <= len(B); j++ {
+			if (A[i - 1] == B[j - 1]) {
+				dp[i][j] = dp[i - 1][j - 1] + 1
+			} else {
+				dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
+			}
+		}
+	}
+	return dp[m][n]
+
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
+
+
+
 JavaScript：
 
 ```javascript

problems/二叉树中递归带着回溯.md

@@ -439,9 +439,84 @@ func traversal(root *TreeNode,result *[]string,path *[]int){
 }
 ```
 
+JavaScript：
 
+100.相同的树
+```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} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+var isSameTree = function(p, q) {
+    if (p === null && q === null) {
+        return true;
+    } else if (p === null || q === null) {
+        return false;
+    } else if (p.val !== q.val) {
+        return false;
+    } else {
+        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+    }
+};
+```
 
+257.二叉树的不同路径
 
+> 回溯法：
+
+```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
+ * @return {string[]}
+ */
+var binaryTreePaths = function(root) {
+    const getPath = (root, path, result) => {
+        path.push(root.val);
+        if (root.left === null && root.right === null) {
+            let n = path.length;
+            let str = '';
+            for (let i=0; i<n-1; i++) {
+                str += path[i] + '->';
+            }
+            str += path[n-1];
+            result.push(str);
+        }
+
+        if (root.left !== null) {
+            getPath(root.left, path, result);
+            path.pop();   // 回溯
+        }
+
+        if (root.right !== null) {
+            getPath(root.right, path, result);
+            path.pop(); 
+        }
+    }
+    
+    if (root === null) return [];
+    let result = [];
+    let path = [];
+    getPath(root, path, result);
+    return result;
+};
+```
 
 
 -----------------------