Leetcode question brushing: String 02 (reverse string II)

541. Reverse string II

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Given a string s And an integer k, You need to check every 2k Before characters k Characters to reverse .

If the remaining characters are less than k individual , Reverse all remaining characters .

If the remaining characters are less than 2k But greater than or equal to k individual , Before reversal k Characters , The rest of the characters remain the same .

Example :

Input : s = “abcdefg”, k = 2
Output : “bacdfeg”

It's every 2*k One for a group （ That's important ！）, In the complete group , front k A reversal , after k No change . The rest is less than half , Namely k, Then reverse all and add , If more than half , Then that k Just reverse it .
There are two special cases , There is less than one group . I took it out alone .

package com.programmercarl.string;
/** * @ClassName ReverseStr * @Descriotion TODO * @Author nitaotao * @Date 2022/6/23 17:45 * @Version 1.0 * https://leetcode.cn/problems/reverse-string-ii/ * 541.  Reverse string  II **/
public class ReverseStr {
    
    public static void main(String[] args) {
    
        System.out.println(reverseStr("abcdefg", 3));
    }

    public static String reverseStr(String s, int k) {
    
        // To put it bluntly, every  [0,k-1]  reverse  [k,2k-1] unchanged 
        //  rotary time Time  45
        int time = s.length() / (2 * k);
        String result = "";

        if (s.length() < k) {
    
            result += reverseString(s.substring(0, s.length()).toCharArray());
            return result;
        }

        if (s.length() <= 2 * k) {
    
            result += reverseString(s.substring(0, k).toCharArray());
            result += s.substring(k);
            return result;
        }

        int index = 0;
        while (index < time) {
    
            result += reverseString(s.substring(index * 2 * k, index * 2 * k + k).toCharArray());
            result += s.substring(index * 2 * k + k, index * 2 * k + 2 * k);
            index++;
        }
        //  More than 
        // If the remaining characters are less than  k  individual , Reverse all remaining characters .
        System.out.println(result);
        if (s.length() - time * 2 * k < k) {
    
            result += reverseString(s.substring(time * 2 * k).toCharArray());
        } else {
    
            // If the remaining characters are less than  2k  But greater than or equal to  k  individual , Before reversal  k  Characters , The rest of the characters remain the same .
            result += reverseString(s.substring(time * 2 * k, time * 2 * k + k).toCharArray());
            result += s.substring(time * 2 * k + k);
        }
        return result;
    }

    public static String reverseString(char[] s) {
    
        // Double finger needling 
        int left = 0;
        int right = s.length - 1;
        while (left < right) {
    
            char temp = s[left];
            s[left] = s[right];
            s[right] = temp;
            left++;
            right--;
        }
        return String.valueOf(s);
    }
}