From ce2c6fb562ae22b2b18230d8f00202f67c89c696 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Fri, 20 Aug 2021 02:55:31 +0800
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补充使用双指针的方法，但是更推荐栈的做法
---
 ...除字符串中的所有相邻重复项.md | 35 +++++++++++++++----
 1 file changed, 29 insertions(+), 6 deletions(-)

diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md
index c4ae85c9..7b059d40 100644
--- a/problems/1047.删除字符串中的所有相邻重复项.md
+++ b/problems/1047.删除字符串中的所有相邻重复项.md
@@ -197,15 +197,38 @@ class Solution {
 
 Python：
 ```python3
+# 方法一，使用栈，推荐！
 class Solution:
     def removeDuplicates(self, s: str) -> str:
-        t = list()
-        for i in s:
-            if t and t[-1] == i:
-                t.pop(-1)
+        res = list()
+        for item in s:
+            if res and res[-1] == item:
+                t.pop()
             else:
-                t.append(i)
-        return "".join(t)  # 字符串拼接
+                t.append(item)
+        return "".join(res)  # 字符串拼接
+```
+
+```python3
+# 双指针
+class Solution:
+    def removeDuplicates(self, s: str) -> str:
+        res = list(s)
+        slow = fast = 0
+        length = len(res)
+
+        while fast < length:
+            # 如果一样直接换，不一样会把后面的填在slow的位置
+            res[slow] = res[fast]
+            
+            # 如果发现和前一个一样，就退一格指针
+            if slow > 0 and res[slow] == res[slow - 1]:
+                slow -= 1
+            else:
+                slow += 1
+            fast += 1
+            
+        return ''.join(res[0: slow])
 ```
 
 Go：