Merge branch 'youngyangyang04:master' into master

problems/0027.移除元素.md

@@ -339,7 +339,6 @@ int removeElement(int* nums, int numsSize, int val){
 }
 ```
 
-
 Kotlin:
 ```kotlin
 fun removeElement(nums: IntArray, `val`: Int): Int {
@@ -351,7 +350,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int {
     }
 ```
 
-
 Scala:
 ```scala
 object Solution {
@@ -368,5 +366,20 @@ object Solution {
 }
 ```
 
+C#:
+```csharp
+public class Solution {
+    public int RemoveElement(int[] nums, int val) {
+        int slow = 0;
+        for (int fast = 0; fast < nums.Length; fast++) {
+            if (val != nums[fast]) {
+                nums[slow++] = nums[fast];
+            }
+        }
+        return slow;
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -480,6 +480,62 @@ var searchRange = function(nums, target) {
     return [-1, -1];
 };
 ```
+
+
+### TypeScript
+
+```typescript
+function searchRange(nums: number[], target: number): number[] {
+    const leftBoard: number = getLeftBorder(nums, target);
+    const rightBoard: number = getRightBorder(nums, target);
+    // target 在nums区间左侧或右侧
+    if (leftBoard === (nums.length - 1) || rightBoard === 0) return [-1, -1];
+    // target 不存在与nums范围内
+    if (rightBoard - leftBoard <= 1) return [-1, -1];
+    // target 存在于nums范围内
+    return [leftBoard + 1, rightBoard - 1];
+};
+// 查找第一个大于target的元素下标
+function getRightBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // 0表示target在nums区间的左边
+    let rightBoard: number = 0;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] <= target) {
+            // 右边界一定在mid右边（不含mid）
+            left = mid + 1;
+            rightBoard = left;
+        } else {
+            // 右边界在mid左边（含mid）
+            right = mid - 1;
+        }
+    }
+    return rightBoard;
+}
+// 查找第一个小于target的元素下标
+function getLeftBorder(nums: number[], target: number): number {
+    let left: number = 0,
+        right: number = nums.length - 1;
+    // length-1表示target在nums区间的右边
+    let leftBoard: number = nums.length - 1;
+    while (left <= right) {
+        let mid = Math.floor((left + right) / 2);
+        if (nums[mid] >= target) {
+            // 左边界一定在mid左边（不含mid）
+            right = mid - 1;
+            leftBoard = right;
+        } else {
+            // 左边界在mid右边（含mid）
+            left = mid + 1;
+        }
+    }
+    return leftBoard;
+}
+```
+
+
 ### Scala
 ```scala
 object Solution {
@@ -527,5 +583,6 @@ object Solution {
 }
 ```
 
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0077.组合.md

@@ -673,5 +673,63 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
 }
 ```
 
+### Scala
+
+暴力:
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+    
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      for (i <- startIndex to n) { // 遍历从startIndex到n
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+    
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
+剪枝:
+
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      // 剪枝优化
+      for (i <- startIndex to (n - (k - path.size) + 1)) { 
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0077.组合优化.md

@@ -346,5 +346,34 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
 }
 ```
 
+Scala:
+
+```scala
+object Solution {
+  import scala.collection.mutable // 导包
+  def combine(n: Int, k: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]() // 存放结果集
+    var path = mutable.ListBuffer[Int]() //存放符合条件的结果
+
+    def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
+      if (path.size == k) {
+        // 如果path的size == k就达到题目要求，添加到结果集，并返回
+        result.append(path.toList)
+        return
+      }
+      // 剪枝优化
+      for (i <- startIndex to (n - (k - path.size) + 1)) { 
+        path.append(i) // 先把数字添加进去
+        backtracking(n, k, i + 1) // 进行下一步回溯
+        path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
+      }
+    }
+
+    backtracking(n, k, 1) // 执行回溯
+    result.toList // 最终返回result的List形式，return关键字可以省略
+  }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0108.将有序数组转换为二叉搜索树.md

@@ -448,5 +448,27 @@ struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
 }
 ```
 
+## Scala
+
+递归:
+
+```scala
+object Solution {
+  def sortedArrayToBST(nums: Array[Int]): TreeNode = {
+    def buildTree(left: Int, right: Int): TreeNode = {
+      if (left > right) return null // 当left大于right的时候，返回空
+      // 最中间的节点是当前节点
+      var mid = left + (right - left) / 2
+      var curNode = new TreeNode(nums(mid))
+      curNode.left = buildTree(left, mid - 1)
+      curNode.right = buildTree(mid + 1, right)
+      curNode
+    }
+    buildTree(0, nums.size - 1)
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0150.逆波兰表达式求值.md

@@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
     return stack.last!
 }
 ```
+
+
+PHP：
+```php
+class Solution {
+    function evalRPN($tokens) {
+        $st = new SplStack();
+        for($i = 0;$i<count($tokens);$i++){
+            // 是数字直接入栈
+            if(is_numeric($tokens[$i])){ 
+                $st->push($tokens[$i]);
+            }else{ 
+                // 是符号进行运算
+                $num1 = $st->pop();
+                $num2 = $st->pop();
+                if ($tokens[$i] == "+") $st->push($num2 + $num1);
+                if ($tokens[$i] == "-") $st->push($num2 - $num1);
+                if ($tokens[$i] == "*") $st->push($num2 * $num1);
+                // 注意处理小数部分
+                if ($tokens[$i] == "/") $st->push(intval($num2 / $num1));
+            }
+        }
+        return $st->pop();
+    }
+}
+```
+
 Scala:
 ```scala
 object Solution {
@@ -351,6 +378,7 @@ object Solution {
     // 最后返回栈顶，不需要加return关键字
     stack.pop() 
   }
+
 }
 ```
 -----------------------

problems/0235.二叉搜索树的最近公共祖先.md

@@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
 
+递归:
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // scala中每个关键字都有其返回值，于是可以不写return
+    if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q)
+    else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q)
+    else root
+  }
+}
+```
+
+迭代:
+
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    var curNode = root  // 因为root是不可变量，所以要赋值给curNode一个可变量
+    while(curNode != null){
+      if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left
+      else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right
+      else return curNode
+    }
+    null
+  }
+}
+```
 
 
 -----------------------

problems/0236.二叉树的最近公共祖先.md

@@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
 };
 ```
 
+## Scala
 
+```scala
+object Solution {
+  def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
+    // 递归结束条件
+    if (root == null || root == p || root == q) {
+      return root
+    }
+
+    var left = lowestCommonAncestor(root.left, p, q)
+    var right = lowestCommonAncestor(root.right, p, q)
+
+    if (left != null && right != null) return root
+    if (left == null) return right
+    left
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0239.滑动窗口最大值.md

@@ -654,8 +654,7 @@ object Solution {
     // 最终返回res，return关键字可以省略
     res
   }
-
-}
+ }
 
 class MyQueue {
   var queue = ArrayBuffer[Int]()
@@ -678,5 +677,84 @@ class MyQueue {
   def peek(): Int = queue.head
 }
 ```
+
+
+PHP:
+```php
+class Solution {
+    /**
+     * @param Integer[] $nums
+     * @param Integer $k
+     * @return Integer[]
+     */
+    function maxSlidingWindow($nums, $k) {
+        $myQueue = new MyQueue();
+        // 先将前k的元素放进队列
+        for ($i = 0; $i < $k; $i++) { 
+            $myQueue->push($nums[$i]);
+        }
+
+        $result = [];
+        $result[] = $myQueue->max(); // result 记录前k的元素的最大值
+
+        for ($i = $k; $i < count($nums); $i++) {
+            $myQueue->pop($nums[$i - $k]); // 滑动窗口移除最前面元素
+            $myQueue->push($nums[$i]); // 滑动窗口前加入最后面的元素
+            $result[]= $myQueue->max(); // 记录对应的最大值
+        }
+        return $result;
+    }
+
+}
+
+// 单调对列构建
+class MyQueue{
+    private $queue;
+
+    public function __construct(){
+        $this->queue = new SplQueue(); //底层是双向链表实现。
+    }
+
+    public function pop($v){
+        // 判断当前对列是否为空
+        // 比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。
+        // bottom 从链表前端查看元素, dequeue 从双向链表的开头移动一个节点
+        if(!$this->queue->isEmpty() && $v == $this->queue->bottom()){
+            $this->queue->dequeue(); //弹出队列
+        }
+    }
+
+    public function push($v){
+        // 判断当前对列是否为空
+        // 如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。
+        // 这样就保持了队列里的数值是单调从大到小的了。
+        while (!$this->queue->isEmpty() && $v > $this->queue->top()) {
+            $this->queue->pop(); // pop从链表末尾弹出一个元素，
+        }
+        $this->queue->enqueue($v);
+    }
+
+    // 查询当前队列里的最大值 直接返回队首
+    public function max(){
+        // bottom 从链表前端查看元素, top从链表末尾查看元素
+        return $this->queue->bottom(); 
+    }
+
+    // 辅助理解: 打印队列元素
+    public function println(){
+        // "迭代器移动到链表头部": 可理解为从头遍历链表元素做准备。
+        // 【PHP中没有指针概念，所以就没说指针。从数据结构上理解，就是把指针指向链表头部】
+        $this->queue->rewind();
+
+        echo "Println: ";
+        while($this->queue->valid()){
+            echo $this->queue->current()," -> ";
+            $this->queue->next();
+        }
+        echo "\n";
+    } 
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0450.删除二叉搜索树中的节点.md

@@ -582,7 +582,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
 };
 ```
 
+## Scala
 
+```scala
+object Solution {
+  def deleteNode(root: TreeNode, key: Int): TreeNode = {
+    if (root == null) return root // 第一种情况，没找到删除的节点，遍历到空节点直接返回
+    if (root.value == key) {
+      // 第二种情况: 左右孩子都为空，直接删除节点，返回null
+      if (root.left == null && root.right == null) return null
+      // 第三种情况: 左孩子为空，右孩子不为空，右孩子补位
+      else if (root.left == null && root.right != null) return root.right
+      // 第四种情况: 左孩子不为空，右孩子为空，左孩子补位
+      else if (root.left != null && root.right == null) return root.left
+      // 第五种情况: 左右孩子都不为空，将删除节点的左子树头节点(左孩子)放到
+      //           右子树的最左边节点的左孩子上，返回删除节点的右孩子
+      else {
+        var tmp = root.right
+        while (tmp.left != null) tmp = tmp.left
+        tmp.left = root.left
+        return root.right
+      }
+    }
+    if (root.value > key) root.left = deleteNode(root.left, key)
+    if (root.value < key) root.right = deleteNode(root.right, key)
+
+    root // 返回根节点，return关键字可以省略
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0538.把二叉搜索树转换为累加树.md

@@ -352,6 +352,24 @@ function convertBST(root: TreeNode | null): TreeNode | null {
 };
 ```
 
+## Scala
+
+```scala
+object Solution {
+  def convertBST(root: TreeNode): TreeNode = {
+    var sum = 0
+    def convert(node: TreeNode): Unit = {
+      if (node == null) return
+      convert(node.right)
+      sum += node.value
+      node.value = sum
+      convert(node.left)
+    }
+    convert(root)
+    root
+  }
+}
+```
 
 
 -----------------------

problems/0669.修剪二叉搜索树.md

@@ -453,7 +453,21 @@ function trimBST(root: TreeNode | null, low: number, high: number): TreeNode | n
 };
 ```
 
+## Scala
 
+递归法:
+```scala
+object Solution {
+  def trimBST(root: TreeNode, low: Int, high: Int): TreeNode = {
+    if (root == null) return null
+    if (root.value < low) return trimBST(root.right, low, high)
+    if (root.value > high) return trimBST(root.left, low, high)
+    root.left = trimBST(root.left, low, high)
+    root.right = trimBST(root.right, low, high)
+    root
+  }
+}
+```
 
 
 -----------------------

problems/0701.二叉搜索树中的插入操作.md

@@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
 ```
 
 
+## Scala
+
+递归:
+
+```scala
+object Solution {
+  def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
+    if (root == null) return new TreeNode(`val`)
+    if (`val` < root.value) root.left = insertIntoBST(root.left, `val`) 
+    else root.right = insertIntoBST(root.right, `val`)
+    root  // 返回根节点
+  }
+}
+```
+
+迭代:
+
+```scala
+object Solution {
+  def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
+    if (root == null) {
+      return new TreeNode(`val`)
+    }
+    var parent = root   // 记录当前节点的父节点
+    var curNode = root
+    while (curNode != null) {
+      parent = curNode
+      if(`val` < curNode.value) curNode = curNode.left
+      else curNode = curNode.right
+    }
+    if(`val` < parent.value) parent.left = new TreeNode(`val`)
+    else parent.right = new TreeNode(`val`)
+    root  // 最终返回根节点
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0704.二分查找.md

@@ -613,6 +613,36 @@ public class Solution{
 }
 ```
 
+**Kotlin:**
+```kotlin
+class Solution {
+    fun search(nums: IntArray, target: Int): Int {
+	// leftBorder
+        var left:Int = 0
+	// rightBorder
+        var right:Int = nums.size - 1
+	// 使用左闭右闭区间
+        while (left <= right) {
+            var middle:Int = left + (right - left)/2
+	    // taget 在左边
+            if (nums[middle] > target) {
+                right = middle - 1
+            }
+            else {
+		// target 在右边
+                if (nums[middle] < target) {
+                    left = middle + 1
+                }
+		// 找到了，返回
+                else   return middle
+                }
+            }
+	    // 没找到，返回
+            return -1   
+        }
+}
+```
+
 
 
 **Kotlin:**

problems/0922.按奇偶排序数组II.md

@@ -260,6 +260,75 @@ var sortArrayByParityII = function(nums) {
 };
 ```
 
+### TypeScript
+
+> 方法一：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const evenArr: number[] = [],
+        oddArr: number[] = [];
+    for (let num of nums) {
+        if (num % 2 === 0) {
+            evenArr.push(num);
+        } else {
+            oddArr.push(num);
+        }
+    }
+    const resArr: number[] = [];
+    for (let i = 0, length = nums.length / 2; i < length; i++) {
+        resArr.push(evenArr[i]);
+        resArr.push(oddArr[i]);
+    }
+    return resArr;
+};
+```
+
+> 方法二：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const length: number = nums.length;
+    const resArr: number[] = [];
+    let evenIndex: number = 0,
+        oddIndex: number = 1;
+    for (let i = 0; i < length; i++) {
+        if (nums[i] % 2 === 0) {
+            resArr[evenIndex] = nums[i];
+            evenIndex += 2;
+        } else {
+            resArr[oddIndex] = nums[i];
+            oddIndex += 2;
+        }
+    }
+    return resArr;
+};
+```
+
+> 方法三：
+
+```typescript
+function sortArrayByParityII(nums: number[]): number[] {
+    const length: number = nums.length;
+    let oddIndex: number = 1;
+    for (let evenIndex = 0; evenIndex < length; evenIndex += 2) {
+        if (nums[evenIndex] % 2 === 1) {
+            // 在偶数位遇到了奇数
+            while (oddIndex < length && nums[oddIndex] % 2 === 1) {
+                oddIndex += 2;
+            }
+            // 在奇数位遇到了偶数，交换
+            let temp = nums[evenIndex];
+            nums[evenIndex] = nums[oddIndex];
+            nums[oddIndex] = temp;
+        }
+    }
+    return nums;
+};
+```
+
+
+
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0977.有序数组的平方.md

@@ -420,6 +420,24 @@ object Solution {
 }
 ```
 
-
+C#：
+```csharp
+public class Solution {
+    public int[] SortedSquares(int[] nums) {
+        int k = nums.Length - 1;
+        int[] result = new int[nums.Length];
+        for (int i = 0, j = nums.Length - 1;i <= j;){
+            if (nums[i] * nums[i] < nums[j] * nums[j]) {
+                result[k--] = nums[j] * nums[j];
+                j--;
+            } else {
+                result[k--] = nums[i] * nums[i];
+                i++;
+            }
+        }
+        return result;
+    }
+}
+```
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>