541. Reverse String II

541. Reverse String II

• Description

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = "abcdefg", k = 2Output: "bacdfeg"

Restrictions:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

class Solution {    public String reverseStr(String s, int k) {             char[] chars=s.toCharArray();            int time=s.length()/(2*k);            System.out.println(time);            char temp;            if(time>0){                for(int i=0;i<time;i++){//change ki ki+1 ki+2....ki+k-1                    for(int x=0;x<(k/2);x++)                    {                        temp=chars[2*k*i+x];                        chars[2*k*i+x]=chars[2*k*i+k-1-x];                        chars[2*k*i+k-1-x]=temp;                    }                }            }            //change 2k*time-1 2k*time...            int left=s.length()-2*k*time;//剩下的数组个数            //2k*time-1 2k*time .....2k*time+left-1            if(left<k){                for(int x=0;x<(left/2);x++){                    temp=chars[2*k*time+x];                    chars[2*k*time+x]=chars[2*k*time-1+left-x];                    chars[2*k*time-1+left-x]=temp;                }            }else{//left>k                for(int x=0;x<(k/2);x++){                    temp=chars[2*k*time+x];                    chars[2*k*time+x]=chars[2*k*time-1+k-x];                    chars[2*k*time-1+k-x]=temp;                }            }           return new String(chars);    }}

improve

class Solution { public String reverseStr(String s, int k) {            char[] chars=s.toCharArray();            int time=s.length()/(2*k);            if(time>0){                for(int i=0;i<time;i++){                        chars=sort(chars,2*k*i,2*k*i+k-1);                  }                   }            int left=s.length()-2*k*time;            int e=left<k?left:k;            if(e>0){                 chars=sort(chars,2*k*time,2*k*time+e-1);            }             return new String(chars);        }    public char[] sort(char[] chars,int s,int e){        //对chars[],start和end之间倒序        int l=e-s+1;        char temp;        for(int i=0;i<(l/2);i++){            temp=chars[s+i];            chars[s+i]=chars[e-i];            chars[e-i]=temp;        }        return chars;     }}