From 6255eae04cefc676d243bc53ea4070bc900b3c99 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Thu, 19 Aug 2021 21:01:20 +0800
Subject: [PATCH] =?UTF-8?q?=E8=A1=A5=E5=85=85=20=E5=89=91=E6=8C=87Offer58-?=
 =?UTF-8?q?II.=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20?=
 =?UTF-8?q?python=E6=96=B9=E6=B3=95?=
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如果不让使用自带函数reversed() 可以使用该方法
---
 problems/剑指Offer58-II.左旋转字符串.md | 17 ++++++++++++++++-
 1 file changed, 16 insertions(+), 1 deletion(-)

diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md
index 1073dafa..6070f85b 100644
--- a/problems/剑指Offer58-II.左旋转字符串.md
+++ b/problems/剑指Offer58-II.左旋转字符串.md
@@ -136,13 +136,28 @@ class Solution:
 
 #         return "".join(s)
 
+# 方法三：如果连reversed也不让使用，那么自己手写一个
+class Solution:
+    def reverseLeftWords(self, s: str, n: int) -> str:
+        def reverse_sub(lst, left, right):
+            while left < right:
+                lst[left], lst[right] = lst[right], lst[left]
+                left += 1
+                right -= 1
+        
+        res = list(s)
+        end = len(res) - 1
+        reverse_sub(res, 0, n - 1)
+        reverse_sub(res, n, end)
+        reverse_sub(res, 0, end)
+        return ''.join(res)
 
 # 时间复杂度：O(n)
 # 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改
 ```
 
 ```python 3
-#方法三：考虑不能用切片的情况下，利用模+下标实现
+#方法四：考虑不能用切片的情况下，利用模+下标实现
 class Solution:
     def reverseLeftWords(self, s: str, n: int) -> str:
         new_s = ''