From fdf0185e1d53e8cda56a6cb040da5b6218fbb3c3 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= <sunxiuyang04@gmail.com>
Date: Tue, 26 Jul 2022 09:17:40 +0800
Subject: [PATCH] =?UTF-8?q?Update=200541.=E5=8F=8D=E8=BD=AC=E5=AD=97?=
 =?UTF-8?q?=E7=AC=A6=E4=B8=B2II.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0541.反转字符串II.md | 34 ++++++++++++++++--------------
 1 file changed, 18 insertions(+), 16 deletions(-)

diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index d9b9466c..8c654a17 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -64,22 +64,7 @@ public:
 ```
 
 
-```
-class Solution {
-public:
-    string reverseStr(string s, int k) {
-        int n=s.size(),pos=0;
-        while(pos<n){
-            //剩余字符串大于等于k的情况
-            if(pos+k<n) reverse(s.begin()+pos,s.begin()+pos+k);
-            //剩余字符串不足k的情况 
-            else reverse(s.begin()+pos,s.end());
-            pos+=2*k;
-        }
-        return s;
-    }
-};
-```
+
 
 那么我们也可以实现自己的reverse函数，其实和题目[344. 反转字符串](https://programmercarl.com/0344.反转字符串.html)道理是一样的。
 
@@ -110,7 +95,24 @@ public:
 ```
 
 
+另一种思路的解法
 
+```CPP
+class Solution {
+public:
+    string reverseStr(string s, int k) {
+        int n = s.size(),pos = 0;
+        while(pos < n){
+            //剩余字符串大于等于k的情况
+            if(pos + k < n) reverse(s.begin() + pos, s.begin() + pos + k);
+            //剩余字符串不足k的情况 
+            else reverse(s.begin() + pos,s.end());
+            pos += 2 * k;
+        }
+        return s;
+    }
+};
+```