力扣题目链接
基础知识

1.解法一

class Solution {
    public String reverseStr(String s, int k) {
        StringBuffer res = new StringBuffer();
        int length = s.length();
        int start = 0;
        while(start<length){

            //找到k处和2k处
            StringBuffer temp = new StringBuffer();
            //与length进行判断，如果大于length,就将其置为length
            int firstK = (start+k>length) ? length : start + k;
            int secondK = (start+2*k>length) ? length : start +2*k;

            //无论start在什么位置，至少反转一次
            temp.append(s.substring(start,firstK));
            res.append(temp.reverse());

            //如果firstK与secondK之间有元素，这些元素可以直接存入res
            if(firstK<secondK){
                res.append(s.substring(firstK,secondK));
            }
            start += (2*k);
        }
        return res.toString();
    }
}

2.解法二

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);
    }
}