---
title: 'LeetCode 541. Reverse String II'
disqus: hackmd
---

# LeetCode 541. Reverse String II

## Description

Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.

If there are fewer 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 leave the other as original.

## Example

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

Input: s = "abcd", k = 2
Output: "bacd"

## Constraints

1 <= s.length <= 10^4^
s consists of only lowercase English letters.
1 <= k <= 10^4^

## Answer

此題因為每隔2k個要轉前k個字元，剩下不足k個的就全轉。一開始先計算長度，若字串長度>k表示能轉k個字元，若長度>2k就往下移動2k，若不夠就往下移動整個剩餘的長度，最後若剩不到k個就再轉剩下的長度即可。

//2022_03_13
void myrev(char *s, int n);
char * reverseStr(char * s, int k){
    char *opr = s;
    int len = 0;
    while(*opr != '\0'){
        len = strlen(opr);
        if(len > k){
            myrev(opr,k);
            opr = len > 2*k ? opr + 2*k : opr + len;
        }
        else{
            myrev(opr,len);
            opr += len;
        }
    }
    return s;
}

void myrev(char *s, int n){
    char *R = s + n - 1;
    while(s<R){
        char tmp = *s;
        *s = *R;
        *R = tmp;
        s++;
        R--;
    }
}

## Link
https://leetcode.com/problems/reverse-string-ii/
###### tags: `Leetcode`