Merge branch 'youngyangyang04:master' into master

problems/0027.移除元素.md

@@ -216,6 +216,25 @@ fn main() {
     println!("{:?}",remove_element(&mut nums, 5));
 }
 ```
+
+Swift:
+
+```swift
+func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
+    var slowIndex = 0
+
+    for fastIndex in 0..<nums.count {
+        if val != nums[fastIndex] {
+            if slowIndex != fastIndex {
+                nums[slowIndex] = nums[fastIndex]
+            }
+            slowIndex += 1
+        }
+    }
+    return slowIndex
+}
+```
+
 -----------------------
 * 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/0035.搜索插入位置.md

@@ -234,8 +234,6 @@ class Solution {
 ```
 
 
-
-
 Python：
 ```python3
 class Solution:
@@ -254,9 +252,6 @@ class Solution:
         return right + 1
 ```
 
-
-Go：
-
 JavaScript:
 ```js
 var searchInsert = function (nums, target) {
@@ -277,6 +272,42 @@ var searchInsert = function (nums, target) {
 };
 ```
 
+Swift:
+
+```swift
+// 暴力法
+func searchInsert(_ nums: [Int], _ target: Int) -> Int {
+    for i in 0..<nums.count {
+        if nums[i] >= target {
+            return i
+        }
+    }
+    return nums.count
+}
+
+// 二分法
+func searchInsert(_ nums: [Int], _ target: Int) -> Int {
+    var left = 0
+    var right = nums.count - 1
+
+    while left <= right {
+        let middle = left + ((right - left) >> 1)
+
+        if nums[middle] > target {
+            right = middle - 1
+        }else if nums[middle] < target {
+            left = middle + 1
+        }else if nums[middle] == target {
+            return middle
+        }
+    }
+
+    return right + 1
+}
+```
+
+
+
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0236.二叉树的最近公共祖先.md

@@ -249,7 +249,6 @@ class Solution {
 ```java
 // 代码精简版
 class Solution {
-    TreeNode pre;
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         if (root == null || root.val == p.val ||root.val == q.val) return root;
         TreeNode left = lowestCommonAncestor(root.left,p,q);

problems/0406.根据身高重建队列.md

@@ -217,7 +217,25 @@ class Solution:
 ```
 
 Go：
-
+```golang
+func reconstructQueue(people [][]int) [][]int {
+    //先将身高从大到小排序，确定最大个子的相对位置
+    sort.Slice(people,func(i,j int)bool{
+        if people[i][0]==people[j][0]{
+            return people[i][1]<people[j][1]//这个才是当身高相同时，将K按照从小到大排序
+        }
+        return people[i][0]>people[j][0]//这个只是确保身高按照由大到小的顺序来排，并不确定K是按照从小到大排序的
+    })
+    //再按照K进行插入排序，优先插入K小的
+    result := make([][]int, 0)
+	for _, info := range people {
+		result = append(result, info)
+		copy(result[info[1] +1:], result[info[1]:])//将插入位置之后的元素后移动一位（意思是腾出空间）
+		result[info[1]] = info//将插入元素位置插入元素
+	}
+	return result
+}
+```
 Javascript:
 ```Javascript
 var reconstructQueue = function(people) {

problems/0513.找树左下角的值.md

@@ -9,6 +9,8 @@
 
 ## 513.找树左下角的值
 
+题目地址：[https://leetcode-cn.com/problems/find-bottom-left-tree-value/](https://leetcode-cn.com/problems/find-bottom-left-tree-value/v)
+
 给定一个二叉树，在树的最后一行找到最左边的值。
 
 示例 1:

problems/0617.合并二叉树.md

@@ -426,6 +426,46 @@ func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
     root1.Right = mergeTrees(root1.Right, root2.Right)
     return root1
 }
+
+// 迭代版本
+func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
+    queue := make([]*TreeNode,0)
+    if root1 == nil{
+        return root2
+    }
+    if root2 == nil{
+        return root1
+    }
+    queue = append(queue,root1)
+    queue = append(queue,root2)
+
+    for size:=len(queue);size>0;size=len(queue){
+        node1 := queue[0]
+        queue = queue[1:]
+        node2 := queue[0]
+        queue = queue[1:]
+        node1.Val += node2.Val
+        // 左子树都不为空
+        if node1.Left != nil && node2.Left != nil{
+            queue = append(queue,node1.Left)
+            queue = append(queue,node2.Left)
+        }
+        // 右子树都不为空
+        if node1.Right !=nil && node2.Right !=nil{
+            queue = append(queue,node1.Right)
+            queue = append(queue,node2.Right)
+        }
+        // 树 1 的左子树为 nil，直接接上树 2 的左子树
+        if node1.Left == nil{
+            node1.Left = node2.Left
+        }
+        // 树 1 的右子树为 nil，直接接上树 2 的右子树
+        if node1.Right == nil{
+            node1.Right = node2.Right
+        }
+    }
+    return root1
+}
 ```
 
 JavaScript: