diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index 03c58b43..9d687cfb 100644
--- a/problems/0027.移除元素.md
+++ b/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;
+ }
+}
+```
+
-----------------------
diff --git a/problems/0077.组合.md b/problems/0077.组合.md
index 8d22d018..fc72be15 100644
--- a/problems/0077.组合.md
+++ b/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
+ }
+}
+```
+
-----------------------
diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md
index a6767047..8d742566 100644
--- a/problems/0077.组合优化.md
+++ b/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
+ }
+}
+```
+
-----------------------
diff --git a/problems/0108.将有序数组转换为二叉搜索树.md b/problems/0108.将有序数组转换为二叉搜索树.md
index c9c1a693..b5f322f0 100644
--- a/problems/0108.将有序数组转换为二叉搜索树.md
+++ b/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)
+ }
+}
+```
+
+
-----------------------
diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md
index 05916135..1a90265a 100644
--- a/problems/0150.逆波兰表达式求值.md
+++ b/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;$ipush($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()
}
+
}
```
-----------------------
diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md
index 9ff7e293..ee86d02f 100644
--- a/problems/0235.二叉搜索树的最近公共祖先.md
+++ b/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
+ }
+}
+```
-----------------------
diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md
index 23695b11..c3e2ae7a 100644
--- a/problems/0236.二叉树的最近公共祖先.md
+++ b/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
+ }
+}
+```
-----------------------
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index 23e79c80..7ee1fdb1 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/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";
+ }
+}
+```
+
-----------------------
diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md
index aca9084f..3fa2a1c5 100644
--- a/problems/0450.删除二叉搜索树中的节点.md
+++ b/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
+ }
+}
+```
-----------------------
diff --git a/problems/0538.把二叉搜索树转换为累加树.md b/problems/0538.把二叉搜索树转换为累加树.md
index 853cca6f..5c1e9e8c 100644
--- a/problems/0538.把二叉搜索树转换为累加树.md
+++ b/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
+ }
+}
+```
-----------------------
diff --git a/problems/0669.修剪二叉搜索树.md b/problems/0669.修剪二叉搜索树.md
index 154ba5a9..a286315d 100644
--- a/problems/0669.修剪二叉搜索树.md
+++ b/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
+ }
+}
+```
-----------------------
diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md
index 102f091e..06e1c88f 100644
--- a/problems/0701.二叉搜索树中的插入操作.md
+++ b/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
+ }
+}
+```
+
-----------------------
diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md
index 6a37e4d1..f3b9326d 100644
--- a/problems/0704.二分查找.md
+++ b/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:**
diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md
index 8ca46db9..49547a15 100644
--- a/problems/0922.按奇偶排序数组II.md
+++ b/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;
+};
+```
+
+
+
-----------------------
diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md
index 4052c570..458107dd 100644
--- a/problems/0977.有序数组的平方.md
+++ b/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;
+ }
+}
+```
-----------------------