From 78b32e8c6c2c7708e1407c729ceebfeeca033f4d Mon Sep 17 00:00:00 2001
From: KailokFung <jialuo_feng@foxmail.com>
Date: Thu, 17 Jun 2021 17:53:56 +0800
Subject: [PATCH] =?UTF-8?q?feat(0541):=20=E5=A2=9E=E5=8A=A0=E5=8F=A6?=
 =?UTF-8?q?=E4=B8=80=E7=A7=8DJava=E7=89=88=E6=9C=AC=E5=86=99=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0541.反转字符串II.md | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index 00581fc0..4b3ed43b 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -103,6 +103,7 @@ public:
 
 Java：
 ```Java
+//解法一
 class Solution {
     public String reverseStr(String s, int k) {
         StringBuffer res = new StringBuffer();
@@ -128,6 +129,28 @@ class Solution {
         return res.toString();
     }
 }
+
+//解法二（似乎更容易理解点）
+//题目的意思其实概括为 每隔2k个反转前k个，尾数不够k个时候全部反转
+class Solution {
+    public String reverseStr(String s, int k) {
+        char[] ch = s.toCharArray();
+        for(int i = 0; i < ch.length; i += 2 * k){
+            int start = i;
+            //这里是判断尾数够不够k个来取决end指针的位置
+            int end = Math.min(ch.length - 1, start + k - 1);
+            //用异或运算反转 
+            while(start < end){
+                ch[start] ^= ch[end];
+                ch[end] ^= ch[start];
+                ch[start] ^= ch[end];
+                start++;
+                end--;
+            }
+        }
+        return new String(ch);
+    }
+}
 ```
 
 Python：