leecode541反转字符串（带有调试代码）

完整带有调试代码

package heima_study.day3;

import java.util.Scanner;

public class 反转字符串541 {
    public static void main(String[] args) {
        Scanner input= new Scanner(System.in);
        String s = input.next();
        int k = input.nextInt();
        String temp = reverseStr(s, k);
        System.out.println(temp);
        input.close();
    }


    public static String reverseStr(String s, int k) {
        char[] ch = s.toCharArray();
        for(int i = 0;i < ch.length;i += 2 * k){
            int start = i;
            // 判断尾数够不够k个来取决end指针的位置
            //这个判断好6啊,本来的if判断直接缩成一句了
            int end = Math.min(ch.length - 1,start + k - 1);
            while(start < end){
                
                char temp = ch[start];
                ch[start] = ch[end];
                ch[end] = temp;

                start++;
                end--;
            }
        }
        return new String(ch);
    }
}

错误代码(不用看,理解错题目意思了)

/*     public static String reverseStr(String s, int k) {
        char[] ch = s.toCharArray();
        //System.out.println(ch);
        //这个题典型的双指针的做法
        int low=0;
        int fast = k*2-1;
        int length = s.length();
        int end=fast;
        while(low<fast&&fast<length){
            //0-2k-1 2k-4k-1; 相差2k-
            char temp  = ch[fast];
            ch[fast]= ch[low];
            ch[low]= temp;
            low++;
            fast--;
            //对于题目的情况,我们直接进行if语句进行判断就行
            //剩余字符刚好2k个
            if((low>=fast)&&end+2*k<length){
                fast=end+2*k;
                low=end+1;
                end=fast;
            }
            // k到2k个
            else if(low>=fast&&end+2*k>length&&end+k<length){
                fast=end+k;
                low=end+1;
                end=fast;                
            }
            //小于k个
            else if(low>=fast&&end+k>length){
                fast=length-1;
                low=end+1;
            }
            //剩余字符刚好大于k小于2k个
        }

        return new String(ch);
    } */