Merge branch 'youngyangyang04:master' into master

Author: RyouMon

README.md

@@ -1,4 +1,4 @@
-👉 推荐 [在线阅读](http://programmercarl.com/) (Github在国内访问经常不稳定)
+👉 推荐 [在线阅读](http://programmercarl.com/) (Github在国内访问经常不稳定)         
 👉 推荐 [Gitee同步](https://gitee.com/programmercarl/leetcode-master) 
 
 > 1. **介绍**：本项目是一套完整的刷题计划，旨在帮助大家少走弯路，循序渐进学算法，[关注作者](#关于作者)
@@ -494,6 +494,7 @@
 ## 图论
 * [463.岛屿的周长](./problems/0463.岛屿的周长.md) （模拟）
 * [841.钥匙和房间](./problems/0841.钥匙和房间.md) 【有向图】dfs，bfs都可以
+* [127.单词接龙](./problems/0127.单词接龙.md) 广搜
 
 ## 并查集 
 * [684.冗余连接](./problems/0684.冗余连接.md) 【并查集基础题目】

problems/0001.两数之和.md

@@ -206,6 +206,23 @@ function twoSum(array $nums, int $target): array
 }
 ```
 
+Swift：
+```swift
+func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
+    var res = [Int]()
+    var dict = [Int : Int]()
+    for i in 0 ..< nums.count {
+        let other = target - nums[i]
+        if dict.keys.contains(other) {
+            res.append(i)
+            res.append(dict[other]!)
+            return res
+        }
+        dict[nums[i]] = i
+    }
+    return res
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0027.移除元素.md

@@ -214,19 +214,17 @@ end
 ```
 Rust:
 ```rust
-pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> &mut Vec<i32> {
-    let mut start: usize = 0;
-    while start < nums.len() {
-        if nums[start] == val {
-            nums.remove(start);
+impl Solution {
+    pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> i32 {
+        let mut slowIdx = 0;
+        for pos in (0..nums.len()) {
+            if nums[pos]!=val {
+                nums[slowIdx] = nums[pos];
+                slowIdx += 1;
+            }
         }
-        start += 1;
+        return (slowIdx) as i32;
     }
-    nums
-}
-fn main() {
-    let mut nums = vec![5,1,3,5,2,3,4,1];
-    println!("{:?}",remove_element(&mut nums, 5));
 }
 ```
 
@@ -248,6 +246,22 @@ func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
 }
 ```
 
+C:
+```c
+int removeElement(int* nums, int numsSize, int val){
+    int slow = 0;
+    for(int fast = 0; fast < numsSize; fast++) {
+        //若快指针位置的元素不等于要删除的元素
+        if(nums[fast] != val) {
+            //将其挪到慢指针指向的位置，慢指针+1
+            nums[slow++] = nums[fast];
+        } 
+    }
+    //最后慢指针的大小就是新的数组的大小
+    return slow;
+}
+```
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)

problems/0046.全排列.md

@@ -183,6 +183,32 @@ class Solution {
     }
 }
 ```
+```java
+// 解法2：通过判断path中是否存在数字，排除已经选择的数字
+class Solution {
+    List<List<Integer>> result = new ArrayList<>();
+    LinkedList<Integer> path = new LinkedList<>();
+    public List<List<Integer>> permute(int[] nums) {
+        if (nums.length == 0) return result;
+        backtrack(nums, path);
+        return result;
+    }
+    public void backtrack(int[] nums, LinkedList<Integer> path) {
+        if (path.size() == nums.length) {
+            result.add(new ArrayList<>(path));
+        }
+        for (int i =0; i < nums.length; i++) {
+            // 如果path中已有，则跳过
+            if (path.contains(nums[i])) {
+                continue;
+            } 
+            path.add(nums[i]);
+            backtrack(nums, path);
+            path.removeLast();
+        }
+    }
+}
+```
 
 Python：
 ```python3

problems/0051.N皇后.md

@@ -341,7 +341,7 @@ class Solution {
 
     public boolean isValid(int row, int col, int n, char[][] chessboard) {
         // 检查列
-        for (int i=0; i<n; ++i) {
+        for (int i=0; i<row; ++i) { // 相当于剪枝
             if (chessboard[i][col] == 'Q') {
                 return false;
             }

problems/0056.合并区间.md

@@ -157,6 +157,28 @@ class Solution {
     }
 }
 ```
+```java
+// 版本2
+class Solution {
+    public int[][] merge(int[][] intervals) {
+        LinkedList<int[]> res = new LinkedList<>();
+        Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));
+        res.add(intervals[0]);
+        for (int i = 1; i < intervals.length; i++) {
+            if (intervals[i][0] <= res.getLast()[1]) {
+                int start = res.getLast()[0];
+                int end = Math.max(intervals[i][1], res.getLast()[1]);
+                res.removeLast();
+                res.add(new int[]{start, end});
+            }
+            else {
+                res.add(intervals[i]);
+            }         
+        }
+        return res.toArray(new int[res.size()][]);
+    }
+}
+```
 
 Python：
 ```python

problems/0059.螺旋矩阵II.md

@@ -373,6 +373,58 @@ func generateMatrix(_ n: Int) -> [[Int]] {
 }
 ```
 
+Rust:
+
+```rust
+impl Solution {
+    pub fn generate_matrix(n: i32) -> Vec<Vec<i32>> {
+        let mut res = vec![vec![0; n as usize]; n as usize];
+        let (mut startX, mut startY, mut offset): (usize, usize, usize) = (0, 0, 1);
+        let mut loopIdx = n/2;
+        let mid: usize = loopIdx as usize;
+        let mut count = 1;
+        let (mut i, mut j): (usize, usize) = (0, 0);
+        while loopIdx > 0 {
+            i = startX;
+            j = startY;
+            
+            while j < (startY + (n as usize) - offset) {
+                res[i][j] = count; 
+                count += 1;
+                j += 1;
+            }
+            
+            while i < (startX + (n as usize) - offset) {
+                res[i][j] = count; 
+                count += 1;
+                i += 1;
+            }
+            
+            while j > startY {
+                res[i][j] = count;
+                count += 1;
+                j -= 1;
+            }
+            
+            while i > startX {
+                res[i][j] = count;
+                count += 1;
+                i -= 1;
+            }
+            
+            startX += 1;
+            startY += 1;   
+            offset += 2; 
+            loopIdx -= 1;
+        }
+        
+        if(n % 2 == 1) {
+            res[mid][mid] = count;
+        }
+        res
+    }
+}
+```
 
 
 -----------------------

problems/0077.组合.md

@@ -435,6 +435,59 @@ func backtrack(n,k,start int,track []int){
 }
 ```
 
+C:
+```c
+int* path;
+int pathTop;
+int** ans;
+int ansTop;
+
+void backtracking(int n, int k,int startIndex) {
+    //当path中元素个数为k个时，我们需要将path数组放入ans二维数组中
+    if(pathTop == k) {
+        //path数组为我们动态申请，若直接将其地址放入二维数组，path数组中的值会随着我们回溯而逐渐变化
+        //因此创建新的数组存储path中的值
+        int* temp = (int*)malloc(sizeof(int) * k);
+        int i;
+        for(i = 0; i < k; i++) {
+            temp[i] = path[i];
+        }
+        ans[ansTop++] = temp;
+        return ;
+    }
+
+    int j;
+    for(j = startIndex; j <=n ;j++) {
+        //将当前结点放入path数组
+        path[pathTop++] = j;
+        //进行递归
+        backtracking(n, k, j + 1);
+        //进行回溯，将数组最上层结点弹出
+        pathTop--;
+    }
+}
+
+int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
+    //path数组存储符合条件的结果
+    path = (int*)malloc(sizeof(int) * k);
+    //ans二维数组存储符合条件的结果数组的集合。（数组足够大，避免极端情况）
+    ans = (int**)malloc(sizeof(int*) * 10000);
+    pathTop = ansTop = 0;
+
+    //回溯算法
+    backtracking(n, k, 1);
+    //最后的返回大小为ans数组大小
+    *returnSize = ansTop;
+    //returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度（都为k）
+    *returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
+    int i;
+    for(i = 0; i < *returnSize; i++) {
+        (*returnColumnSizes)[i] = k;
+    }
+    //返回ans二维数组
+    return ans;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0104.二叉树的最大深度.md

@@ -311,6 +311,35 @@ class solution {
 }
 ```
 
+### 559.n叉树的最大深度 
+```java
+class solution {
+    /**
+     * 迭代法，使用层序遍历
+     */
+    public int maxDepth(Node root) {
+        if (root == null)   return 0;
+        int depth = 0;
+        Queue<Node> que = new LinkedList<>();
+        que.offer(root);
+        while (!que.isEmpty())
+        {
+            depth ++;
+            int len = que.size();
+            while (len > 0)
+            {
+                Node node = que.poll();
+                for (int i = 0; i < node.children.size(); i++)
+                    if (node.children.get(i) != null) 
+                        que.offer(node.children.get(i));
+                len--;
+            }
+        }
+        return depth;
+    }
+}
+```
+
 ## python 
 
 ### 104.二叉树的最大深度
@@ -505,21 +534,51 @@ var maxdepth = function(root) {
 
 二叉树最大深度层级遍历
 ```javascript
-var maxdepth = function(root) {
-    //使用递归的方法 递归三部曲
-    //1. 确定递归函数的参数和返回值
-    const getdepth=function(node){
-    //2. 确定终止条件
-        if(node===null){
-            return 0;
+var maxDepth = function(root) {
+    if(!root) return 0
+    let count = 0
+    const queue = [root]
+    while(queue.length) {
+        let size = queue.length
+        /* 层数+1 */
+        count++
+        while(size--) {
+            let node = queue.shift();
+            node.left && queue.push(node.left);
+            node.right && queue.push(node.right);
+        }
+    }
+    return count
+};
+```
+
+N叉树的最大深度  递归写法
+```js
+var maxDepth = function(root) {
+    if(!root) return 0
+    let depth = 0
+    for(let node of root.children) {
+        depth = Math.max(depth, maxDepth(node))
+    }
+    return depth + 1
+}
+```
+
+N叉树的最大深度  层序遍历
+```js
+var maxDepth = function(root) {
+    if(!root) return 0
+    let count = 0
+    let queue = [root]
+    while(queue.length) {
+        let size = queue.length
+        count++
+        while(size--) {
+            let node = queue.shift()
+            node && (queue = [...queue, ...node.children])
         }
-    //3. 确定单层逻辑
-        let leftdepth=getdepth(node.left);
-        let rightdepth=getdepth(node.right);
-        let depth=1+math.max(leftdepth,rightdepth);
-        return depth;
     }
-    return getDepth(root);
+    return count
 };
 ```