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：