From f6eee532e49acba882be2b202825ee7082d9a69c Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Thu, 19 Aug 2021 16:26:42 +0800
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另一种解法进行了不必要的操作，且代码可读性较差。
推荐使用第一种解法
---
 problems/0344.反转字符串.md | 17 +++++------------
 1 file changed, 5 insertions(+), 12 deletions(-)

diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md
index 4f96f839..763097be 100644
--- a/problems/0344.反转字符串.md
+++ b/problems/0344.反转字符串.md
@@ -162,21 +162,14 @@ class Solution:
         Do not return anything, modify s in-place instead.
         """
         left, right = 0, len(s) - 1
-        while(left < right):
+        
+        # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低
+        # 推荐该写法，更加通俗易懂
+        while left < right:
             s[left], s[right] = s[right], s[left]
             left += 1
             right -= 1
-            
-# 下面的写法更加简洁，但是都是同样的算法
-# class Solution:
-#     def reverseString(self, s: List[str]) -> None:
-#         """
-#         Do not return anything, modify s in-place instead.
-#         """
-          # 不需要判别是偶数个还是奇数个序列，因为奇数个的时候，中间那个不需要交换就可
-#         for i in range(len(s)//2):
-#             s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i]
-#         return s
+       
 ```
 
 Go：