#Leetcode# 541. Reverse String II

https://leetcode.com/problems/reverse-string-ii/

 

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 = 2
Output: "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]

代码：

class Solution {
public:
    string reverseStr(string s, int k) {
        int len = s.length();
        int cnt = len / k;
        for(int i = 0; i <= cnt; i ++) {
            if(i % 2 == 0) {
                if(i * k + k < len)
                    reverse(s.begin() + i * k, s.begin() + i * k + k);
                else reverse(s.begin() + i * k, s.end());
            }
        }
        return s;
    }
};

　　reverse 函数好像很久不用了呢

reverse 用法：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
string s;
char a[maxn];

int main() {
    getline(cin, s);
    reverse(s.begin(), s.end());
    cout << s;

    printf("\n-----\n");

    gets(a);
    int len = strlen(a);
    reverse(a, a + len);
    printf("%s", a);
    return 0;
}