Merge branch 'youngyangyang04:master' into master

problems/0017.电话号码的字母组合.md

@@ -557,6 +557,37 @@ func letterCombinations(_ digits: String) -> [String] {
 }
 ```
 
+## Scala:
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def letterCombinations(digits: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if(digits == "") return result.toList // 如果参数为空，返回空结果集的List形式
+    var path = mutable.ListBuffer[Char]()
+    // 数字和字符的映射关系
+    val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
+
+    def backtracking(index: Int): Unit = {
+      if (index == digits.size) {
+        result.append(path.mkString)  // mkString语法：将数组类型直接转换为字符串
+        return
+      }
+      var digit = digits(index) - '0' // 这里使用toInt会报错！必须 -'0'
+      for (i <- 0 until map(digit).size) {
+        path.append(map(digit)(i))
+        backtracking(index + 1)
+        path = path.take(path.size - 1)
+      }
+    }
+    
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0024.两两交换链表中的节点.md

@@ -254,20 +254,19 @@ TypeScript：
 
 ```typescript
 function swapPairs(head: ListNode | null): ListNode | null {
-  const dummyHead: ListNode = new ListNode(0, head);
+    const dummyNode: ListNode = new ListNode(0, head);
-  let cur: ListNode = dummyHead;
+    let curNode: ListNode | null = dummyNode;
-  while(cur.next !== null && cur.next.next !== null) {
+    while (curNode && curNode.next && curNode.next.next) {
-    const tem: ListNode = cur.next;
+        let firstNode: ListNode = curNode.next,
-    const tem1: ListNode = cur.next.next.next;
+            secNode: ListNode = curNode.next.next,
-    
+            thirdNode: ListNode | null = curNode.next.next.next;
-    cur.next = cur.next.next; // step 1
+        curNode.next = secNode;
-    cur.next.next = tem; // step 2
+        secNode.next = firstNode;
-    cur.next.next.next = tem1; // step 3
+        firstNode.next = thirdNode;
-    
+        curNode = firstNode;
-    cur = cur.next.next;
     }
-  return dummyHead.next;
+    return dummyNode.next;
-}
+};
 ```
 
 Kotlin:

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/0035.搜索插入位置.md

@@ -283,6 +283,28 @@ var searchInsert = function (nums, target) {
 };
 ```
 
+### TypeScript
+
+```typescript
+// 第一种二分法
+function searchInsert(nums: number[], target: number): number {
+    const length: number = nums.length;
+    let left: number = 0,
+        right: number = length - 1;
+    while (left <= right) {
+        const mid: number = Math.floor((left + right) / 2);
+        if (nums[mid] < target) {
+            left = mid + 1;
+        } else if (nums[mid] === target) {
+            return mid;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return right + 1;
+};
+```
+
 ### Swift 
 
 ```swift

problems/0039.组合总和.md

@@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(sum: Int, index: Int): Unit = {
+      if (sum == target) {
+        result.append(path.toList) // 如果正好等于target，就添加到结果集
+        return
+      }
+      // 应该是从当前索引开始的，而不是从0
+      // 剪枝优化：添加循环守卫，当sum + c(i) <= target的时候才循环，才可以进入下一次递归
+      for (i <- index until candidates.size if sum + candidates(i) <= target) {
+        path.append(candidates(i))
+        backtracking(sum + candidates(i), i)
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(0, 0)
+    result.toList
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0040.组合总和II.md

@@ -693,5 +693,37 @@ func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] {
 }
 ```
 
+
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable    
+  def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = {
+    var res = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+    var candidate = candidates.sorted
+
+    def backtracking(sum: Int, startIndex: Int): Unit = {
+      if (sum == target) {
+        res.append(path.toList)
+        return
+      }
+
+      for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
+        if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
+          path.append(candidate(i))
+          backtracking(sum + candidate(i), i + 1)
+          path = path.take(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0, 0)
+    res.toList
+  }
+}
+```
+
 -----------------------
 <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/0093.复原IP地址.md

@@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def restoreIpAddresses(s: String): List[String] = {
+    var result = mutable.ListBuffer[String]()
+    if (s.size < 4 || s.length > 12) return result.toList
+    var path = mutable.ListBuffer[String]()
+
+    // 判断IP中的一个字段是否为正确的
+    def isIP(sub: String): Boolean = {
+      if (sub.size > 1 && sub(0) == '0') return false
+      if (sub.toInt > 255) return false
+      true
+    }
+
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        if (path.size == 4) {
+          result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接
+          return
+        }
+        return
+      }
+      // subString
+      for (i <- startIndex until startIndex + 3 if i < s.size) {
+        var subString = s.substring(startIndex, i + 1)
+        if (isIP(subString)) { // 如果合法则进行下一轮
+          path.append(subString)
+          backtracking(i + 1)
+          path = path.take(path.size - 1)
+        }
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。
 
 ```java
 class Solution {
+    Map<Integer, Integer> map;  // 方便根据数值查找位置
     public TreeNode buildTree(int[] inorder, int[] postorder) {
-        return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
+        map = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+            map.put(inorder[i], i);
         }
-    public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
+
-                               int[] postorder, int postLeft, int postRight) {
+        return findNode(inorder,  0, inorder.length, postorder,0, postorder.length);  // 前闭后开
-        // 没有元素了
+    }
-        if (inRight - inLeft < 1) {
+    
+    public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
+        // 参数里的范围都是前闭后开
+        if (inBegin >= inEnd || postBegin >= postEnd) {  // 不满足左闭右开，说明没有元素，返回空树
             return null;
         }
-        // 只有一个元素了
+        int rootIndex = map.get(postorder[postEnd - 1]);  // 找到后序遍历的最后一个元素在中序遍历中的位置
-        if (inRight - inLeft == 1) {
+        TreeNode root = new TreeNode(inorder[rootIndex]);  // 构造结点
-            return new TreeNode(inorder[inLeft]);
+        int lenOfLeft = rootIndex - inBegin;  // 保存中序左子树个数，用来确定后序数列的个数
-        }
+        root.left = findNode(inorder, inBegin, rootIndex,
-        // 后序数组postorder里最后一个即为根结点
+                            postorder, postBegin, postBegin + lenOfLeft);
-        int rootVal = postorder[postRight - 1];
+        root.right = findNode(inorder, rootIndex + 1, inEnd,
-        TreeNode root = new TreeNode(rootVal);
+                            postorder, postBegin + lenOfLeft, postEnd - 1);
-        int rootIndex = 0;
+
-        // 根据根结点的值找到该值在中序数组inorder里的位置
-        for (int i = inLeft; i < inRight; i++) {
-            if (inorder[i] == rootVal) {
-                rootIndex = i;
-                break;
-            }
-        }
-        // 根据rootIndex划分左右子树
-        root.left = buildTree1(inorder, inLeft, rootIndex,
-                postorder, postLeft, postLeft + (rootIndex - inLeft));
-        root.right = buildTree1(inorder, rootIndex + 1, inRight,
-                postorder, postLeft + (rootIndex - inLeft), postRight - 1);
         return root;
     }
 }
@@ -622,31 +616,29 @@ class Solution {
 
 ```java
 class Solution {
+    Map<Integer, Integer> map;
     public TreeNode buildTree(int[] preorder, int[] inorder) {
-        return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
+        map = new HashMap<>();
+        for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+            map.put(inorder[i], i);
         }
 
-    public TreeNode helper(int[] preorder, int preLeft, int preRight,
+        return findNode(preorder, 0, preorder.length, inorder,  0, inorder.length);  // 前闭后开
-                           int[] inorder, int inLeft, int inRight) {
-        // 递归终止条件
-        if (inLeft > inRight || preLeft > preRight) return null;
-
-        // val 为前序遍历第一个的值，也即是根节点的值
-        // idx 为根据根节点的值来找中序遍历的下标
-        int idx = inLeft, val = preorder[preLeft];
-        TreeNode root = new TreeNode(val);
-        for (int i = inLeft; i <= inRight; i++) {
-            if (inorder[i] == val) {
-                idx = i;
-                break;
-            }
     }
 
-        // 根据 idx 来递归找左右子树
+    public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
-        root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
+        // 参数里的范围都是前闭后开
-                         inorder, inLeft, idx - 1);
+        if (preBegin >= preEnd || inBegin >= inEnd) {  // 不满足左闭右开，说明没有元素，返回空树
-        root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
+            return null;
-                         inorder, idx + 1, inRight);
+        }
+        int rootIndex = map.get(preorder[preBegin]);  // 找到前序遍历的第一个元素在中序遍历中的位置
+        TreeNode root = new TreeNode(inorder[rootIndex]);  // 构造结点
+        int lenOfLeft = rootIndex - inBegin;  // 保存中序左子树个数，用来确定前序数列的个数
+        root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,  
+                            inorder, inBegin, rootIndex);
+        root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
+                            inorder, rootIndex + 1, inEnd);
+
         return root;
     }
 }

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/0131.分割回文串.md

@@ -676,5 +676,50 @@ impl Solution {
     }
 }
 ```
+
+
+## Scala
+
+```scala
+object Solution {
+
+  import scala.collection.mutable
+
+  def partition(s: String): List[List[String]] = {
+    var result = mutable.ListBuffer[List[String]]()
+    var path = mutable.ListBuffer[String]()
+
+    // 判断字符串是否回文
+    def isPalindrome(start: Int, end: Int): Boolean = {
+      var (left, right) = (start, end)
+      while (left < right) {
+        if (s(left) != s(right)) return false
+        left += 1
+        right -= 1
+      }
+      true
+    }
+
+    // 回溯算法
+    def backtracking(startIndex: Int): Unit = {
+      if (startIndex >= s.size) {
+        result.append(path.toList)
+        return
+      }
+      // 添加循环守卫，如果当前分割是回文子串则进入回溯
+      for (i <- startIndex until s.size if isPalindrome(startIndex, i)) {
+        path.append(s.substring(startIndex, i + 1))
+        backtracking(i + 1)
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(0)
+    result.toList
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0141.环形链表.md

@@ -7,6 +7,8 @@
 
 # 141. 环形链表
 
+[力扣题目链接](https://leetcode.cn/problems/linked-list-cycle/submissions/)
+
 给定一个链表，判断链表中是否有环。
 
 如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。
@@ -103,7 +105,7 @@ class Solution:
         return False
 ```
 
-## Go
+### Go
 
 ```go
 func hasCycle(head *ListNode) bool {
@@ -139,6 +141,23 @@ var hasCycle = function(head) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function hasCycle(head: ListNode | null): boolean {
+    let slowNode: ListNode | null = head,
+        fastNode: ListNode | null = head;
+    while (fastNode !== null && fastNode.next !== null) {
+        slowNode = slowNode!.next;
+        fastNode = fastNode.next.next;
+        if (slowNode === fastNode) return true;
+    }
+    return false;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0142.环形链表II.md

@@ -301,13 +301,13 @@ function detectCycle(head: ListNode | null): ListNode | null {
     let slowNode: ListNode | null = head,
         fastNode: ListNode | null = head;
     while (fastNode !== null && fastNode.next !== null) {
-        slowNode = (slowNode as ListNode).next;
+        slowNode = slowNode!.next;
         fastNode = fastNode.next.next;
         if (slowNode === fastNode) {
             slowNode = head;
             while (slowNode !== fastNode) {
-                slowNode = (slowNode as ListNode).next;
+                slowNode = slowNode!.next;
-                fastNode = (fastNode as ListNode).next;
+                fastNode = fastNode!.next;
             }
             return slowNode;
         }

problems/0143.重排链表.md

@@ -6,6 +6,8 @@
 
 # 143.重排链表
 
+[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
+
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png)
 
 ## 思路
@@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) {
 }
 ```
 
+### TypeScript
+
+> 辅助数组法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null) return;
+    const helperArr: ListNode[] = [];
+    let curNode: ListNode | null = head;
+    while (curNode !== null) {
+        helperArr.push(curNode);
+        curNode = curNode.next;
+    }
+    let node: ListNode = head;
+    let left: number = 1,
+        right: number = helperArr.length - 1;
+    let count: number = 0;
+    while (left <= right) {
+        if (count % 2 === 0) {
+            node.next = helperArr[right--];
+        } else {
+            node.next = helperArr[left++];
+        }
+        count++;
+        node = node.next;
+    }
+    node.next = null;
+};
+```
+
+> 分割链表法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null || head.next === null) return;
+    let fastNode: ListNode = head,
+        slowNode: ListNode = head;
+    while (fastNode.next !== null && fastNode.next.next !== null) {
+        slowNode = slowNode.next!;
+        fastNode = fastNode.next.next;
+    }
+    let head1: ListNode | null = head;
+    // 反转后半部分链表
+    let head2: ListNode | null = reverseList(slowNode.next);
+    // 分割链表
+    slowNode.next = null;
+    /**
+    	直接在head1链表上进行插入
+        head1 链表长度一定大于或等于head2,
+        因此在下面的循环中，只要head2不为null, head1 一定不为null
+     */
+    while (head2 !== null) {
+        const tempNode1: ListNode | null = head1!.next,
+            tempNode2: ListNode | null = head2.next;
+        head1!.next = head2;
+        head2.next = tempNode1;
+        head1 = tempNode1;
+        head2 = tempNode2;
+    }
+};
+function reverseList(head: ListNode | null): ListNode | null {
+    let curNode: ListNode | null = head,
+        preNode: ListNode | null = null;
+    while (curNode !== null) {
+        const tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+}
+```
+
 ### C
+
 方法三：反转链表
 ```c
 //翻转链表

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/0209.长度最小的子数组.md

@@ -465,6 +465,27 @@ object Solution {
   }
 }
 ```
-
+C#:
+```csharp
+public class Solution {
+    public int MinSubArrayLen(int s, int[] nums) {
+        int n = nums.Length;
+        int ans = int.MaxValue;
+        int start = 0, end = 0;
+        int sum = 0;
+        while (end < n)  {
+            sum += nums[end];
+            while (sum >= s) 
+            {
+                ans = Math.Min(ans, end - start + 1);
+                sum -= nums[start];
+                start++;
+            }
+            end++;
+        }
+        return ans == int.MaxValue ? 0 : ans;
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0216.组合总和III.md

@@ -502,5 +502,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] {
 }
 ```
 
+## Scala
+
+```scala
+object Solution {
+  import scala.collection.mutable
+  def combinationSum3(k: Int, n: Int): List[List[Int]] = {
+    var result = mutable.ListBuffer[List[Int]]()
+    var path = mutable.ListBuffer[Int]()
+
+    def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = {
+      if (sum > n) return // 剪枝，如果sum>目标和，就返回
+      if (sum == n && path.size == k) {
+        result.append(path.toList)
+        return
+      }
+      // 剪枝
+      for (i <- startIndex to (9 - (k - path.size) + 1)) {
+        path.append(i)
+        backtracking(k, n, sum + i, i + 1) 
+        path = path.take(path.size - 1)
+      }
+    }
+
+    backtracking(k, n, 0, 1) // 调用递归方法
+    result.toList // 最终返回结果集的List形式
+  }
+}
+```
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0234.回文链表.md

@@ -273,7 +273,7 @@ class Solution:
         return pre
 ```
 
-## Go
+### Go
 
 ```go
 
@@ -319,6 +319,63 @@ var isPalindrome = function(head) {
 };
 ```
 
+### TypeScript
+
+> 数组模拟
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+    const helperArr: number[] = [];
+    let curNode: ListNode | null = head;
+    while (curNode !== null) {
+        helperArr.push(curNode.val);
+        curNode = curNode.next;
+    }
+    let left: number = 0,
+        right: number = helperArr.length - 1;
+    while (left < right) {
+        if (helperArr[left++] !== helperArr[right--]) return false;
+    }
+    return true;
+};
+```
+
+> 反转后半部分链表
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+    if (head === null || head.next === null) return true;
+    let fastNode: ListNode | null = head,
+        slowNode: ListNode = head,
+        preNode: ListNode = head;
+    while (fastNode !== null && fastNode.next !== null) {
+        preNode = slowNode;
+        slowNode = slowNode.next!;
+        fastNode = fastNode.next.next;
+    }
+    preNode.next = null;
+    let cur1: ListNode | null = head;
+    let cur2: ListNode | null = reverseList(slowNode);
+    while (cur1 !== null) {
+        if (cur1.val !== cur2!.val) return false;
+        cur1 = cur1.next;
+        cur2 = cur2!.next;
+    }
+    return true;
+};
+function reverseList(head: ListNode | null): ListNode | null {
+    let curNode: ListNode | null = head,
+        preNode: ListNode | null = null;
+    while (curNode !== null) {
+        let tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+}
+```
+
 
 
 -----------------------

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/0344.反转字符串.md

@@ -190,13 +190,13 @@ javaScript:
  * @return {void} Do not return anything, modify s in-place instead.
  */
 var reverseString = function(s) {
-    return s.reverse();
+    //Do not return anything, modify s in-place instead.
+    reverse(s)
 };
 
-var reverseString = function(s) {
+var reverse = function(s) {
     let l = -1, r = s.length;
     while(++l < --r) [s[l], s[r]] = [s[r], s[l]];
-    return s;
 };
 ```
 

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/0707.设计链表.md

@@ -104,8 +104,9 @@ public:
     // 在第index个节点之前插入一个新节点，例如index为0，那么新插入的节点为链表的新头节点。
     // 如果index 等于链表的长度，则说明是新插入的节点为链表的尾结点
     // 如果index大于链表的长度，则返回空
+    // 如果index小于0，则置为0，作为链表的新头节点。
     void addAtIndex(int index, int val) {
-        if (index > _size) {
+        if (index > _size || index < 0) {
             return;
         }
         LinkedNode* newNode = new LinkedNode(val);

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;
+};
+```
+
+
+
 
 -----------------------
 <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

@@ -421,6 +421,24 @@ 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;
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/背包理论基础01背包-1.md

@@ -356,10 +356,14 @@ func test_2_wei_bag_problem1(weight, value []int, bagweight int) int {
 	// 递推公式
 	for i := 1; i < len(weight); i++ {
 		//正序,也可以倒序
-		for  j := weight[i];j<= bagweight ; j++ {
+	    for j := 0; j <= bagweight; j++ {
+	        if j < weight[i] {
+	            dp[i][j] = dp[i-1][j]
+	        } else {
 	            dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
 	        }
 	    }		
+	}
 	return dp[len(weight)-1][bagweight]
 }