diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
index 2fcfd283..c7a2cfcb 100644
--- a/problems/0019.删除链表的倒数第N个节点.md
+++ b/problems/0019.删除链表的倒数第N个节点.md
@@ -165,20 +165,17 @@ class Solution:
  * }
  */
 func removeNthFromEnd(head *ListNode, n int) *ListNode {
-    dummyHead := &ListNode{}
-    dummyHead.Next = head
-    cur := head
-    prev := dummyHead
-    i := 1
-    for cur != nil {
-        cur = cur.Next
-        if i > n {
-            prev = prev.Next
-        }
-        i++
-    }
-    prev.Next = prev.Next.Next
-    return dummyHead.Next
+	dummyNode := &ListNode{0, head}
+	fast, slow := dummyNode, dummyNode
+	for i := 0; i <= n; i++ { // 注意<=，否则快指针为空时，慢指针正好在倒数第n个上面
+		fast = fast.Next
+	}
+	for fast != nil {
+		fast = fast.Next
+		slow = slow.Next
+	}
+	slow.Next = slow.Next.Next
+	return dummyNode.Next
 }
 ```
 
diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index 3aadb67e..daf8dac3 100644
--- a/problems/0027.移除元素.md
+++ b/problems/0027.移除元素.md
@@ -224,6 +224,27 @@ class Solution:
                 
 ```
 
+``` python 3
+# 相向双指针法
+# 时间复杂度 O(n)
+# 空间复杂度 O(1)
+class Solution:
+    def removeElement(self, nums: List[int], val: int) -> int:
+        n = len(nums)
+        left, right  = 0, n - 1
+        while left <= right:
+            while left <= right and nums[left] != val:
+                left += 1
+            while left <= right and nums[right] == val:
+                right -= 1
+            if left < right:
+                nums[left] = nums[right]
+                left += 1
+                right -= 1
+        return left
+                
+```
+
 ### Go：
 
 ```go
diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md
index 4fa294cf..622b1117 100644
--- a/problems/0070.爬楼梯完全背包版本.md
+++ b/problems/0070.爬楼梯完全背包版本.md
@@ -165,7 +165,21 @@ class climbStairs{
 ```
 
 ### Python3：
+```python3
+def climbing_stairs(n,m):
+    dp = [0]*(n+1) # 背包总容量
+    dp[0] = 1 
+    # 排列题，注意循环顺序，背包在外物品在内
+    for j in range(1,n+1):
+        for i in range(1,m+1):
+            if j>=i:
+                dp[j] += dp[j-i] # 这里i就是重量而非index
+    return dp[n]
 
+if __name__ == '__main__':
+    n,m = list(map(int,input().split(' ')))
+    print(climbing_stairs(n,m))
+```
 
 
 ### Go：
diff --git a/problems/0332.重新安排行程.md b/problems/0332.重新安排行程.md
index 036cfc51..6c8a4814 100644
--- a/problems/0332.重新安排行程.md
+++ b/problems/0332.重新安排行程.md
@@ -418,34 +418,7 @@ class Solution {
 }
 ```
 
-### Python 
-回溯 使用used数组 
-
-```python
-class Solution:
-    def findItinerary(self, tickets: List[List[str]]) -> List[str]:
-        tickets.sort() # 先排序，这样一旦找到第一个可行路径，一定是字母排序最小的
-        used = [0] * len(tickets)
-        path = ['JFK']
-        results = []
-        self.backtracking(tickets, used, path, 'JFK', results)
-        return results[0]
-    
-    def backtracking(self, tickets, used, path, cur, results):
-        if len(path) == len(tickets) + 1:  # 终止条件：路径长度等于机票数量+1
-            results.append(path[:])  # 将当前路径添加到结果列表
-            return True
-        
-        for i, ticket in enumerate(tickets):  # 遍历机票列表
-            if ticket[0] == cur and used[i] == 0:  # 找到起始机场为cur且未使用过的机票
-                used[i] = 1  # 标记该机票为已使用
-                path.append(ticket[1])  # 将到达机场添加到路径中
-                state = self.backtracking(tickets, used, path, ticket[1], results)  # 递归搜索
-                path.pop()  # 回溯，移除最后添加的到达机场
-                used[i] = 0  # 标记该机票为未使用
-                if state:
-                    return True  # 只要找到一个可行路径就返回，不继续搜索
-
+### Python
 ```
 回溯 使用字典
 ```python