diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md
index 0beb45cf..06367851 100644
--- a/problems/0046.全排列.md
+++ b/problems/0046.全排列.md
@@ -182,7 +182,45 @@ class Solution {
 ```
 
 Python：
+```python3
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        res = []  #存放符合条件结果的集合
+        path = []  #用来存放符合条件的结果
+        used = []  #用来存放已经用过的数字
+        def backtrack(nums,used):
+            if len(path) == len(nums):
+                return res.append(path[:])  #此时说明找到了一组
+            for i in range(0,len(nums)):
+                if nums[i] in used:
+                    continue  #used里已经收录的元素，直接跳过
+                path.append(nums[i])
+                used.append(nums[i])
+                backtrack(nums,used)
+                used.pop()
+                path.pop()
+        backtrack(nums,used)
+        return res
+```
 
+Python（优化，不用used数组）：
+```python3
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        res = []  #存放符合条件结果的集合
+        path = []  #用来存放符合条件的结果
+        def backtrack(nums):
+            if len(path) == len(nums):
+                return res.append(path[:])  #此时说明找到了一组
+            for i in range(0,len(nums)):
+                if nums[i] in path:  #path里已经收录的元素，直接跳过
+                    continue
+                path.append(nums[i])
+                backtrack(nums)  #递归
+                path.pop()  #回溯
+        backtrack(nums)
+        return res
+```
 
 Go：
 ```Go
diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md
index b4b42a4a..8618515e 100644
--- a/problems/0055.跳跃游戏.md
+++ b/problems/0055.跳跃游戏.md
@@ -141,7 +141,20 @@ func canJUmp(nums []int) bool {
 }
 ```
 
-
+Javascript:
+```Javascript
+var canJump = function(nums) {
+    if(nums.length === 1) return true
+    let cover = 0
+    for(let i = 0; i <= cover; i++) {
+        cover = Math.max(cover, i + nums[i])
+        if(cover >= nums.length - 1) {
+            return true
+        }
+    }
+    return false
+};
+```
 
 
 -----------------------
diff --git a/problems/0078.子集.md b/problems/0078.子集.md
index 17c4fb5c..0b2f3c09 100644
--- a/problems/0078.子集.md
+++ b/problems/0078.子集.md
@@ -205,7 +205,20 @@ class Solution {
 ```
 
 Python：
-
+```python3
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        res = []  
+        path = []  
+        def backtrack(nums,startIndex):
+            res.append(path[:])  #收集子集，要放在终止添加的上面，否则会漏掉自己
+            for i in range(startIndex,len(nums)):  #当startIndex已经大于数组的长度了，就终止了，for循环本来也结束了，所以不需要终止条件
+                path.append(nums[i])
+                backtrack(nums,i+1)  #递归
+                path.pop()  #回溯
+        backtrack(nums,0)
+        return res
+```
 
 Go：
 ```Go
diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md
index 6b12a95b..71aef5c7 100644
--- a/problems/0090.子集II.md
+++ b/problems/0090.子集II.md
@@ -208,7 +208,23 @@ class Solution {
 ```
 
 Python：
-
+```python3
+class Solution:
+    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+        res = []  #存放符合条件结果的集合
+        path = []  #用来存放符合条件结果
+        def backtrack(nums,startIndex):
+            res.append(path[:])
+            for i in range(startIndex,len(nums)):
+                if i > startIndex and nums[i] == nums[i - 1]:  #我们要对同一树层使用过的元素进行跳过
+                    continue
+                path.append(nums[i])
+                backtrack(nums,i+1)  #递归
+                path.pop()  #回溯
+        nums = sorted(nums)  #去重需要排序
+        backtrack(nums,0)
+        return res
+```
 
 Go：
 ```Go
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index 15c2d1f9..9deb1e0c 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -234,25 +234,19 @@ class Solution:
 ```
 
 Go：
-```func detectCycle(head *ListNode) *ListNode {
-    if head ==nil{
-        return head 
-    }
-    slow:=head
-    fast:=head.Next
-
-    for fast!=nil&&fast.Next!=nil{
-        if fast==slow{
-            slow=head
-            fast=fast.Next
-            for fast!=slow {
-                fast=fast.Next
-                slow=slow.Next
+```go
+func detectCycle(head *ListNode) *ListNode {
+    slow, fast := head, head
+    for fast != nil && fast.Next != nil {
+        slow = slow.Next
+        fast = fast.Next.Next
+        if slow == fast {
+            for slow != head {
+                slow = slow.Next
+                head = head.Next
             }
-            return slow
+            return head
         }
-        fast=fast.Next.Next
-        slow=slow.Next
     }
     return nil
 }
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 e38bc1a3..c8b0da08 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
@@ -224,6 +224,22 @@ var evalRPN = function(tokens) {
 };
 ```
 
+python3
+
+```python
+def evalRPN(tokens) -> int:
+    stack = list()
+    for i in range(len(tokens)):
+        if tokens[i] not in ["+", "-", "*", "/"]:
+            stack.append(tokens[i])
+        else:
+            tmp1 = stack.pop()
+            tmp2 = stack.pop()
+            res = eval(tmp2+tokens[i]+tmp1)
+            stack.append(str(int(res)))
+    return stack[-1]
+```
+
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md
index 10314a08..c2af71f7 100644
--- a/problems/0232.用栈实现队列.md
+++ b/problems/0232.用栈实现队列.md
@@ -282,6 +282,50 @@ class MyQueue {
 
 
 Python：
+```python
+# 使用两个栈实现先进先出的队列
+class MyQueue:
+    def __init__(self):
+        """
+        Initialize your data structure here.
+        """
+        self.stack1 = list()
+        self.stack2 = list()
+
+    def push(self, x: int) -> None:
+        """
+        Push element x to the back of queue.
+        """
+        # self.stack1用于接受元素
+        self.stack1.append(x)
+
+    def pop(self) -> int:
+        """
+        Removes the element from in front of queue and returns that element.
+        """
+        # self.stack2用于弹出元素，如果self.stack2为[],则将self.stack1中元素全部弹出给self.stack2
+        if self.stack2 == []:
+            while self.stack1:
+                tmp = self.stack1.pop()
+                self.stack2.append(tmp)
+        return self.stack2.pop()
+
+    def peek(self) -> int:
+        """
+        Get the front element.
+        """
+        if self.stack2 == []:
+            while self.stack1:
+                tmp = self.stack1.pop()
+                self.stack2.append(tmp)
+        return self.stack2[-1]
+
+    def empty(self) -> bool:
+        """
+        Returns whether the queue is empty.
+        """
+        return self.stack1 == [] and self.stack2 == []
+```
 
 
 Go：
diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md
index c9423a1a..fefaa293 100644
--- a/problems/0343.整数拆分.md
+++ b/problems/0343.整数拆分.md
@@ -51,10 +51,6 @@ dp[i]的定义讲贯彻整个解题过程，下面哪一步想不懂了，就想
 
 **那有同学问了，j怎么就不拆分呢？**
 
-j是从1开始遍历，拆分j的情况，在遍历j的过程中其实都计算过了。
-
-**那有同学问了，j怎么就不拆分呢？**
-
 j是从1开始遍历，拆分j的情况，在遍历j的过程中其实都计算过了。那么从1遍历j，比较(i - j) * j和dp[i - j] * j 取最大的。递推公式：dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j));
 
 也可以这么理解，j * (i - j) 是单纯的把整数拆分为两个数相乘，而j * dp[i - j]是拆分成两个以及两个以上的个数相乘。
@@ -213,8 +209,19 @@ class Solution {
 ```
 
 Python：
-
-
+```python
+class Solution:
+    def integerBreak(self, n: int) -> int:
+        dp = [0] * (n + 1)
+        dp[2] = 1
+        for i in range(3, n + 1):
+            # 假设对正整数 i 拆分出的第一个正整数是 j（1 <= j < i），则有以下两种方案：
+            # 1) 将 i 拆分成 j 和 i−j 的和，且 i−j 不再拆分成多个正整数，此时的乘积是 j * (i-j)
+            # 2) 将 i 拆分成 j 和 i−j 的和，且 i−j 继续拆分成多个正整数，此时的乘积是 j * dp[i-j]
+            for j in range(1, i):
+                dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
+        return dp[n]
+```
 Go：
 
 
diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md
index 30d702b4..e5daa9db 100644
--- a/problems/0404.左叶子之和.md
+++ b/problems/0404.左叶子之和.md
@@ -226,7 +226,51 @@ class Solution:
 ```
 Go：
 
-
+JavaScript:
+递归版本
+```javascript
+var sumOfLeftLeaves = function(root) {
+    //采用后序遍历 递归遍历
+    // 1. 确定递归函数参数
+    const nodesSum = function(node){
+        // 2. 确定终止条件
+        if(node===null){
+            return 0;
+        }
+        let leftValue = sumOfLeftLeaves(node.left);
+        let rightValue = sumOfLeftLeaves(node.right);
+        // 3. 单层递归逻辑
+        let midValue = 0;
+        if(node.left&&node.left.left===null&&node.left.right===null){
+            midValue = node.left.val;
+        }
+        let sum = midValue + leftValue + rightValue;
+        return sum;
+    }
+    return nodesSum(root);
+};
+```
+迭代版本
+```javascript
+var sumOfLeftLeaves = function(root) {
+   //采用层序遍历
+   if(root===null){
+       return null;
+   }
+   let queue = [];
+   let sum = 0;
+   queue.push(root);
+   while(queue.length){
+     let node = queue.shift();
+     if(node.left!==null&&node.left.left===null&&node.left.right===null){
+         sum+=node.left.val;
+     }
+     node.left&&queue.push(node.left);
+     node.right&&queue.push(node.right);
+   }
+   return sum;
+};
+```
 
 
 
diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md
index 691f7aef..5538a2c9 100644
--- a/problems/0491.递增子序列.md
+++ b/problems/0491.递增子序列.md
@@ -229,7 +229,28 @@ class Solution {
 
 
 Python：
-
+```python3
+class Solution:
+    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
+        res = []
+        path = []
+        def backtrack(nums,startIndex):
+            repeat = []  #这里使用数组来进行去重操作
+            if len(path) >=2:
+                res.append(path[:])  #注意这里不要加return，要取树上的节点
+            for i in range(startIndex,len(nums)):
+                if nums[i] in repeat:
+                    continue
+                if len(path) >= 1:
+                    if nums[i] < path[-1]:
+                        continue
+                repeat.append(nums[i])  #记录这个元素在本层用过了，本层后面不能再用了
+                path.append(nums[i])
+                backtrack(nums,i+1)
+                path.pop()
+        backtrack(nums,0)
+        return res
+```
 
 Go：
 
diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index db53caf7..6c8f3e94 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -106,27 +106,24 @@ Java：
 class Solution {
     public String reverseStr(String s, int k) {
         StringBuffer res = new StringBuffer();
-
-        for (int i = 0; i < s.length(); i += (2 * k)) {
+        int length = s.length();
+        int start = 0;
+        while (start < length) {
+            // 找到k处和2k处
             StringBuffer temp = new StringBuffer();
-            // 剩余字符大于 k 个，每隔 2k 个字符的前 k 个字符进行反转
-            if (i + k <= s.length()) {
-                // 反转前 k 个字符
-                temp.append(s.substring(i, i + k));
-                res.append(temp.reverse());
+            // 与length进行判断，如果大于length了，那就将其置为length
+            int firstK = (start + k > length) ? length : start + k;
+            int secondK = (start + (2 * k) > length) ? length : start + (2 * k);
 
-                // 反转完前 k 个字符之后，如果紧接着还有 k 个字符，则直接加入这 k 个字符
-                if (i + 2 * k <= s.length()) {
-                    res.append(s.substring(i + k, i + 2 * k));
-                // 不足 k 个字符，则直接加入剩下所有字符
-                } else {
-                    res.append(s.substring(i + k, s.length()));
-                }
-                continue;
-            }
-            // 剩余字符少于 k 个，则将剩余字符全部反转。
-            temp.append(s.substring(i, s.length()));
+            //无论start所处位置，至少会反转一次
+            temp.append(s.substring(start, firstK));
             res.append(temp.reverse());
+
+            // 如果firstK到secondK之间有元素，这些元素直接放入res里即可。
+            if (firstK < secondK) { //此时剩余长度一定大于k。
+                res.append(s.substring(firstK, secondK));
+            }
+            start += (2 * k);
         }
         return res.toString();
     }
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 167a258c..b5d392e6 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -100,10 +100,66 @@ public:
 Java：
 
 Python：
-
+```Python
+class Solution:
+    def sortedSquares(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        i,j,k = 0,n - 1,n - 1
+        ans = [-1] * n
+        while i <= j:
+            lm = nums[i] ** 2
+            rm = nums[j] ** 2
+            if lm > rm:
+                ans[k] = lm
+                i += 1
+            else:
+                ans[k] = rm
+                j -= 1
+            k -= 1
+        return ans
+```
 
 Go：
-
+```Go
+func sortedSquares(nums []int) []int {
+	n := len(nums)
+	i, j, k := 0, n-1, n-1
+	ans := make([]int, n)
+	for i <= j {
+		lm, rm := nums[i]*nums[i], nums[j]*nums[j]
+		if lm > rm {
+			ans[k] = lm
+			i++
+		} else {
+			ans[k] = rm
+			j--
+		}
+		k--
+	}
+	return ans
+}
+```
+Rust
+```
+impl Solution {
+    pub fn sorted_squares(nums: Vec<i32>) -> Vec<i32> {
+        let n = nums.len();
+        let (mut i,mut j,mut k) = (0,n - 1,n- 1);
+        let mut ans = vec![0;n];
+        while i <= j{
+            if nums[i] * nums[i] < nums[j] * nums[j] {
+                ans[k] = nums[j] * nums[j];
+                j -= 1;
+            }else{
+                ans[k] = nums[i] * nums[i];
+                i += 1;
+            }
+            k -= 1;
+        }
+        ans
+    }
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 9e97264c..13014dd1 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -155,6 +155,42 @@ Python：
 
 Go：
 
+```go
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+    curA := headA
+    curB := headB
+    lenA, lenB := 0, 0
+    // 求A，B的长度
+    for curA != nil {
+        curA = curA.Next
+        lenA++
+    }
+    for curB != nil {
+        curB = curB.Next
+        lenB++
+    }
+    var step int
+    var fast, slow *ListNode
+    // 请求长度差，并且让更长的链表先走相差的长度
+    if lenA > lenB {
+        step = lenA - lenB
+        fast, slow = headA, headB
+    } else {
+        step = lenB - lenA
+        fast, slow = headB, headA
+    }
+    for i:=0; i < step; i++ {
+        fast = fast.Next
+    }
+    // 遍历两个链表遇到相同则跳出遍历
+    for fast != slow {
+        fast = fast.Next
+        slow = slow.Next
+    }
+    return fast
+}
+```
+
 javaScript: 
 
 ```js