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feat(0541): 增加另一种Java版本写法

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Original file line number	Diff line number	Diff line change
`@@ -103,6 +103,7 @@ public:`
`103`	`103`
`104`	`104`	`Java:`
`105`	`105`	```Java
	`106`	`+//解法一`
`106`	`107`	`class Solution {`
`107`	`108`	`public String reverseStr(String s, int k) {`
`108`	`109`	`StringBuffer res = new StringBuffer();`
`@@ -128,6 +129,28 @@ class Solution {`
`128`	`129`	`return res.toString();`
`129`	`130`	`}`
`130`	`131`	`}`
	`132`	`+`
	`133`	`+//解法二(似乎更容易理解点)`
	`134`	`+//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`
	`135`	`+class Solution {`
	`136`	`+ public String reverseStr(String s, int k) {`
	`137`	`+ char[] ch = s.toCharArray();`
	`138`	`+ for(int i = 0; i < ch.length; i += 2 * k){`
	`139`	`+ int start = i;`
	`140`	`+ //这里是判断尾数够不够k个来取决end指针的位置`
	`141`	`+ int end = Math.min(ch.length - 1, start + k - 1);`
	`142`	`+ //用异或运算反转`
	`143`	`+ while(start < end){`
	`144`	`+ ch[start] ^= ch[end];`
	`145`	`+ ch[end] ^= ch[start];`
	`146`	`+ ch[start] ^= ch[end];`
	`147`	`+ start++;`
	`148`	`+ end--;`
	`149`	`+ }`
	`150`	`+ }`
	`151`	`+ return new String(ch);`
	`152`	`+ }`
	`153`	`+}`
`131`	`154`	```
`132`	`155`
`133`	`156`	`Python:`

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