leetcode 541. Reverse String II 反转字符串

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

本题题意十分简单，就是做一个字符串的反转，注意是没2k个子串中前k个反转，其余的不变

建议和leetcode 344. Reverse String 反转字符串 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    string reverseStr(string s, int k)    {        int count = s.length() / k;        for (int i = 0; i <= count; i++)        {            if (i % 2 == 0)            {                if (i*k + k < s.length())                    reverse(s.begin()+i*k,s.begin() + i*k+k);                else                    reverse(s.begin() + i*k, s.end());            }        }        return s;    }};