diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md
index 079ca834..b6c12919 100644
--- a/problems/0047.全排列II.md
+++ b/problems/0047.全排列II.md
@@ -5,6 +5,7 @@
   <a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
 </p>
 <p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
+
 # 排列问题（二）
 
 ## 47.全排列 II
@@ -222,6 +223,43 @@ class Solution:
         return res
 ```
 
+Go：
+
+```go
+var res [][]int
+func permute(nums []int) [][]int {
+    res = [][]int{}
+    sort.Ints(nums)
+    dfs(nums, make([]int, 0), make([]bool, len(nums)))
+    return res
+}
+
+func dfs(nums, path []int, used []bool) {
+    if len(path) == len(nums) {
+        res = append(res, append([]int{}, path...))
+        return
+    }
+
+    m := make(map[int]bool)
+    for i := 0; i < len(nums); i++ {
+        // used 从剩余 nums 中选
+        if used[i] {
+            continue
+        }
+        // m 集合间去重
+        if _, ok := m[nums[i]]; ok {
+            continue
+        }
+        m[nums[i]] = true
+        path = append(path, nums[i])
+        used[i] = true
+        dfs(nums, path, used)
+        used[i] = false
+        path = path[:len(path)-1]
+    }
+}
+```
+
 Javascript:
 
 ```javascript
diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md
index a04ebc0c..248d10f1 100644
--- a/problems/0098.验证二叉搜索树.md
+++ b/problems/0098.验证二叉搜索树.md
@@ -344,7 +344,7 @@ Python：
 #         self.val = val
 #         self.left = left
 #         self.right = right
-//递归法
+# 递归法
 class Solution:
     def isValidBST(self, root: TreeNode) -> bool:
         res = []  //把二叉搜索树按中序遍历写成list
@@ -356,6 +356,35 @@ class Solution:
             return res  
         buildalist(root)
         return res == sorted(res) and len(set(res)) == len(res) //检查list里的数有没有重复元素，以及是否按从小到大排列
+
+# 简单递归法
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        def isBST(root, min_val, max_val):
+            if not root: return True
+            if root.val >= max_val or root.val <= min_val:
+                return False
+            return isBST(root.left, min_val, root.val) and isBST(root.right, root.val, max_val)
+        return isBST(root, float("-inf"), float("inf"))
+
+# 迭代-中序遍历
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        stack = []
+        cur = root
+        pre = None
+        while cur or stack:
+            if cur: # 指针来访问节点，访问到最底层
+                stack.append(cur)
+                cur = cur.left
+            else: # 逐一处理节点
+                cur = stack.pop()
+                if pre and cur.val <= pre.val: # 比较当前节点和前节点的值的大小
+                    return False
+                pre = cur 
+                cur = cur.right
+        return True
+
 ```
 Go：
 ```Go
diff --git a/problems/0123.买卖股票的最佳时机III.md b/problems/0123.买卖股票的最佳时机III.md
index fccb187d..7ff1bfe2 100644
--- a/problems/0123.买卖股票的最佳时机III.md
+++ b/problems/0123.买卖股票的最佳时机III.md
@@ -193,35 +193,49 @@ dp[1] = max(dp[1], dp[0] - prices[i]); 如果dp[1]取dp[1]，即保持买入股
 Java：
 
 ```java
-class Solution { // 动态规划
+// 版本一
+class Solution {
     public int maxProfit(int[] prices) {
-        // 可交易次数
-        int k = 2;
+        int len = prices.length;
+        // 边界判断, 题目中 length >= 1, 所以可省去
+        if (prices.length == 0) return 0;
 
-        // [天数][交易次数][是否持有股票] 
-        int[][][] dp = new int[prices.length][k + 1][2];
+        /*
+         * 定义 5 种状态:
+         * 0: 没有操作, 1: 第一次买入, 2: 第一次卖出, 3: 第二次买入, 4: 第二次卖出
+         */
+        int[][] dp = new int[len][5];
+        dp[0][1] = -prices[0];
+        // 初始化第二次买入的状态是确保 最后结果是最多两次买卖的最大利润
+        dp[0][3] = -prices[0];
 
-        // badcase
-        dp[0][0][0] = 0;
-        dp[0][0][1] = Integer.MIN_VALUE;
-        dp[0][1][0] = 0;
-        dp[0][1][1] = -prices[0];
-        dp[0][2][0] = 0;
-        dp[0][2][1] = Integer.MIN_VALUE;
-
-        for (int i = 1; i < prices.length; i++) {
-            for (int j = 2; j >= 1; j--) {
-                // dp公式
-                dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
-                dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
-            }
+        for (int i = 1; i < len; i++) {
+            dp[i][1] = Math.max(dp[i - 1][1], -prices[i]);
+            dp[i][2] = Math.max(dp[i - 1][2], dp[i][1] + prices[i]);
+            dp[i][3] = Math.max(dp[i - 1][3], dp[i][2] - prices[i]);
+            dp[i][4] = Math.max(dp[i - 1][4], dp[i][3] + prices[i]);
         }
 
-        int res = 0;
-        for (int i = 1; i < 3; i++) {
-            res = Math.max(res, dp[prices.length - 1][i][0]);
+        return dp[len - 1][4];
+    }
+}
+
+// 版本二: 空间优化
+class Solution {
+    public int maxProfit(int[] prices) {
+        int len = prices.length;
+        int[] dp = new int[5];
+        dp[1] = -prices[0];
+        dp[3] = -prices[0];
+
+        for (int i = 1; i < len; i++) {
+            dp[1] = Math.max(dp[1], dp[0] - prices[i]);
+            dp[2] = Math.max(dp[2], dp[1] + prices[i]);
+            dp[3] = Math.max(dp[3], dp[2] - prices[i]);
+            dp[4] = Math.max(dp[4], dp[3] + prices[i]);
         }
-        return res;
+
+        return dp[4];
     }
 }
 ```
diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md
index b6a6242e..59892ef9 100644
--- a/problems/0139.单词拆分.md
+++ b/problems/0139.单词拆分.md
@@ -291,6 +291,27 @@ func wordBreak(s string,wordDict []string) bool  {
 }
 ```
 
+Javascript：
+```javascript
+const wordBreak = (s, wordDict) => {
+
+    let dp = Array(s.length + 1).fill(false);
+    dp[0] = true;
+
+    for(let i = 0; i <= s.length; i++){
+        for(let j = 0; j < wordDict.length; j++) {
+            if(i >= wordDict[j].length) {
+                if(s.slice(i - wordDict[j].length, i) === wordDict[j] && dp[i - wordDict[j].length]) {
+                    dp[i] = true
+                }
+            }
+        }
+    }
+
+    return dp[s.length];
+}
+```
+
 
 
 -----------------------
diff --git a/problems/0188.买卖股票的最佳时机IV.md b/problems/0188.买卖股票的最佳时机IV.md
index 431c292b..d1ca8a8d 100644
--- a/problems/0188.买卖股票的最佳时机IV.md
+++ b/problems/0188.买卖股票的最佳时机IV.md
@@ -170,41 +170,54 @@ public:
 Java：
 
 ```java
-class Solution { //动态规划
+// 版本一: 三维 dp数组
+class Solution {
     public int maxProfit(int k, int[] prices) {
-        if (prices == null || prices.length < 2 || k == 0) {
-            return 0;
+        if (prices.length == 0) return 0;
+
+        // [天数][交易次数][是否持有股票]
+        int len = prices.length;
+        int[][][] dp = new int[len][k + 1][2];
+        
+        // dp数组初始化
+        // 初始化所有的交易次数是为确保 最后结果是最多 k 次买卖的最大利润
+        for (int i = 0; i <= k; i++) {
+            dp[0][i][1] = -prices[0];
         }
 
-        // [天数][交易次数][是否持有股票] 
-        int[][][] dp = new int[prices.length][k + 1][2];
-
-        // bad case
-        dp[0][0][0] = 0;
-        dp[0][0][1] = Integer.MIN_VALUE;
-        dp[0][1][0] = 0;
-        dp[0][1][1] = -prices[0];
-        // dp[0][j][0] 都均为0
-        // dp[0][j][1] 异常值都取Integer.MIN_VALUE;
-        for (int i = 2; i < k + 1; i++) {
-            dp[0][i][0] = 0;
-            dp[0][i][1] = Integer.MIN_VALUE;
-        }
-
-        for (int i = 1; i < prices.length; i++) {
-            for (int j = k; j >= 1; j--) {
-                // dp公式
+        for (int i = 1; i < len; i++) {
+            for (int j = 1; j <= k; j++) {
+                // dp方程, 0表示不持有/卖出, 1表示持有/买入
                 dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
                 dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
             }
         }
+        return dp[len - 1][k][0];
+    }
+}
 
-        int res = 0;
-        for (int i = 1; i < k + 1; i++) {
-            res = Math.max(res, dp[prices.length - 1][i][0]);
+// 版本二: 空间优化
+class Solution {
+    public int maxProfit(int k, int[] prices) {
+        if (prices.length == 0) return 0;
+
+        // [天数][股票状态]
+        // 股票状态: 奇数表示第 k 次交易持有/买入, 偶数表示第 k 次交易不持有/卖出, 0 表示没有操作
+        int len = prices.length;
+        int[][] dp = new int[len][k*2 + 1];
+        
+        // dp数组的初始化, 与版本一同理
+        for (int i = 1; i < k*2; i += 2) {
+            dp[0][i] = -prices[0];
         }
 
-        return res;
+        for (int i = 1; i < len; i++) {
+            for (int j = 0; j < k*2 - 1; j += 2) {
+                dp[i][j + 1] = Math.max(dp[i - 1][j + 1], dp[i - 1][j] - prices[i]);
+                dp[i][j + 2] = Math.max(dp[i - 1][j + 2], dp[i - 1][j + 1] + prices[i]);
+            }
+        }
+        return dp[len - 1][k*2];
     }
 }
 ```
diff --git a/problems/0322.零钱兑换.md b/problems/0322.零钱兑换.md
index cf537088..302ae789 100644
--- a/problems/0322.零钱兑换.md
+++ b/problems/0322.零钱兑换.md
@@ -304,6 +304,26 @@ func min(a, b int) int {
 
 ```
 
+Javascript：
+```javascript
+const coinChange = (coins, amount) => {
+    if(!amount) {
+        return 0;
+    }
+
+    let dp = Array(amount + 1).fill(Infinity);
+    dp[0] = 0;
+
+    for(let i =0; i < coins.length; i++) {
+        for(let j = coins[i]; j <= amount; j++) {
+            dp[j] = Math.min(dp[j - coins[i]] + 1, dp[j]);
+        }
+    }
+
+    return dp[amount] === Infinity ? -1 : dp[amount];
+}
+```
+
 
 
 -----------------------
diff --git a/problems/0377.组合总和Ⅳ.md b/problems/0377.组合总和Ⅳ.md
index d0394706..2ee009b3 100644
--- a/problems/0377.组合总和Ⅳ.md
+++ b/problems/0377.组合总和Ⅳ.md
@@ -201,6 +201,25 @@ func combinationSum4(nums []int, target int) int {
 }
 ```
 
+Javascript：
+```javascript
+const combinationSum4 = (nums, target) => {
+
+    let dp = Array(target + 1).fill(0);
+    dp[0] = 1;
+
+    for(let i = 0; i <= target; i++) {
+        for(let j = 0; j < nums.length; j++) {
+            if (i >= nums[j]) {
+                dp[i] += dp[i - nums[j]];
+            }
+        }
+    }
+
+    return dp[target];
+};
+```
+
 
 
 -----------------------
diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index fd925f76..6b73c080 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -244,6 +244,35 @@ func max(a,b int) int {
 }
 ```
 
+Javascript：
+```javascript
+const findMaxForm = (strs, m, n) => {
+    const dp = Array.from(Array(m+1), () => Array(n+1).fill(0));
+    let numOfZeros, numOfOnes;
+
+    for(let str of strs) {
+        numOfZeros = 0;
+        numOfOnes = 0;
+    
+        for(let c of str) {
+            if (c === '0') {
+                numOfZeros++;
+            } else {
+                numOfOnes++;
+            }
+        }
+
+        for(let i = m; i >= numOfZeros; i--) {
+            for(let j = n; j >= numOfOnes; j--) {
+                dp[i][j] = Math.max(dp[i][j], dp[i - numOfZeros][j - numOfOnes] + 1);
+            }
+        }
+    }
+
+    return dp[m][n];
+};
+```
+
 
 
 -----------------------
diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md
index 85c9d307..103b8b10 100644
--- a/problems/0491.递增子序列.md
+++ b/problems/0491.递增子序列.md
@@ -257,6 +257,34 @@ class Solution:
 
 Go：
 
+```golang
+func findSubsequences(nums []int) [][]int {
+    var subRes []int
+    var res [][]int
+    backTring(0,nums,subRes,&res)
+    return res
+}
+func backTring(startIndex int,nums,subRes []int,res *[][]int){
+    if len(subRes)>1{
+    tmp:=make([]int,len(subRes))
+    copy(tmp,subRes)
+    *res=append(*res,tmp)
+    }
+    history:=[201]int{}//记录本层元素使用记录
+    for i:=startIndex;i<len(nums);i++{
+        //分两种情况判断：一，当前取的元素小于子集的最后一个元素，则继续寻找下一个适合的元素
+        //                或者二，当前取的元素在本层已经出现过了，所以跳过该元素，继续寻找
+        if len(subRes)>0&&nums[i]<subRes[len(subRes)-1]||history[nums[i] + 100]==1{
+            continue
+        }
+        history[nums[i] + 100]=1//表示本层该元素使用过了
+        subRes=append(subRes,nums[i])
+        backTring(i+1,nums,subRes,res)
+        subRes=subRes[:len(subRes)-1]
+    }
+}
+```
+
 Javascript:
 
 ```Javascript
diff --git a/problems/0496.下一个更大元素I.md b/problems/0496.下一个更大元素I.md
index 88e89e52..0404e434 100644
--- a/problems/0496.下一个更大元素I.md
+++ b/problems/0496.下一个更大元素I.md
@@ -185,3 +185,26 @@ public:
 
 建议大家把情况一二三想清楚了，先写出版本一的代码，然后在其基础上在做精简！
 
+## 其他语言版本
+
+Python：
+```python3
+class Solution:
+    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        result = [-1]*len(nums1)
+        stack = [0]
+        for i in range(1,len(nums2)):
+            # 情况一情况二
+            if nums2[i]<=nums2[stack[-1]]:
+                stack.append(i)
+            # 情况三
+            else:
+                while len(stack)!=0 and nums2[i]>nums2[stack[-1]]:
+                    if nums2[stack[-1]] in nums1:
+                        index = nums1.index(nums2[stack[-1]])
+                        result[index]=nums2[i]
+                    stack.pop()                 
+                stack.append(i)
+        return result
+```
+
diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md
index 08043ea3..3f907da9 100644
--- a/problems/0518.零钱兑换II.md
+++ b/problems/0518.零钱兑换II.md
@@ -243,6 +243,22 @@ func change(amount int, coins []int) int {
 }
 ```
 
+Javascript：
+```javascript
+const change = (amount, coins) => {
+    let dp = Array(amount + 1).fill(0);
+    dp[0] = 1;
+
+    for(let i =0; i < coins.length; i++) {
+        for(let j = coins[i]; j <= amount; j++) {
+            dp[j] += dp[j - coins[i]];
+        }
+    }
+
+    return dp[amount];
+}
+```
+
 
 
 -----------------------
diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md
index 47b2b434..bf646443 100644
--- a/problems/0530.二叉搜索树的最小绝对差.md
+++ b/problems/0530.二叉搜索树的最小绝对差.md
@@ -222,6 +222,26 @@ class Solution:
         for i in range(len(res)-1):  // 统计有序数组的最小差值
             r = min(abs(res[i]-res[i+1]),r)
         return r
+        
+# 迭代法-中序遍历
+class Solution:
+    def getMinimumDifference(self, root: TreeNode) -> int:
+        stack = []
+        cur = root
+        pre = None
+        result = float('inf')
+        while cur or stack:
+            if cur: # 指针来访问节点，访问到最底层
+                stack.append(cur)
+                cur = cur.left
+            else: # 逐一处理节点
+                cur = stack.pop()
+                if pre: # 当前节点和前节点的值的差值
+                    result = min(result, cur.val - pre.val)
+                pre = cur
+                cur = cur.right
+        return result
+        
 ```
 Go：
 > 中序遍历，然后计算最小差值
diff --git a/problems/0739.每日温度.md b/problems/0739.每日温度.md
index e72fd6a4..8ad79fe3 100644
--- a/problems/0739.每日温度.md
+++ b/problems/0739.每日温度.md
@@ -211,7 +211,24 @@ Java：
     }
 ```
 Python：
-
+``` Python3
+class Solution:
+    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
+        answer = [0]*len(temperatures)
+        stack = [0]
+        for i in range(1,len(temperatures)):
+            # 情况一和情况二
+            if temperatures[i]<=temperatures[stack[-1]]:
+                stack.append(i)
+            # 情况三
+            else:
+                while len(stack) != 0 and temperatures[i]>temperatures[stack[-1]]:
+                    answer[stack[-1]]=i-stack[-1]
+                    stack.pop()
+                stack.append(i)
+            
+        return answer
+```
 Go：
 
 > 暴力法
diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md
index 2ddab584..8e245c20 100644
--- a/problems/1143.最长公共子序列.md
+++ b/problems/1143.最长公共子序列.md
@@ -8,6 +8,8 @@
 
 ## 1143.最长公共子序列 
 
+题目链接: https://leetcode-cn.com/problems/longest-common-subsequence/
+
 给定两个字符串 text1 和 text2，返回这两个字符串的最长公共子序列的长度。  
 
 一个字符串的 子序列 是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。     
diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md
index a5a708cf..1ad09c4b 100644
--- a/problems/背包问题理论基础完全背包.md
+++ b/problems/背包问题理论基础完全背包.md
@@ -176,9 +176,48 @@ int main() {
 
 ## 其他语言版本
 
-
 Java：
 
+```java
+		//先遍历物品，再遍历背包
+    private static void testCompletePack(){
+        int[] weight = {1, 3, 4};
+        int[] value = {15, 20, 30};
+        int bagWeight = 4;
+        int[] dp = new int[bagWeight + 1];
+        for (int i = 0; i < weight.length; i++){
+            for (int j = 1; j <= bagWeight; j++){
+                if (j - weight[i] >= 0){
+                    dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);
+                }
+            }
+        }
+        for (int maxValue : dp){
+            System.out.println(maxValue + "   ");
+        }
+    }
+
+    //先遍历背包，再遍历物品
+    private static void testCompletePackAnotherWay(){
+        int[] weight = {1, 3, 4};
+        int[] value = {15, 20, 30};
+        int bagWeight = 4;
+        int[] dp = new int[bagWeight + 1];
+        for (int i = 1; i <= bagWeight; i++){
+            for (int j = 0; j < weight.length; j++){
+                if (i - weight[j] >= 0){
+                    dp[i] = Math.max(dp[i], dp[i - weight[j]] + value[j]);
+                }
+            }
+        }
+        for (int maxValue : dp){
+            System.out.println(maxValue + "   ");
+        }
+    }
+```
+
+
+
 Python：
 
 ```python3