Reverse String II问题及解法

本文主要是介绍Reverse String II问题及解法，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

问题描述：

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.

示例：

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

问题分析：

题目要求每2k个字符中，翻转前k个，实现起来比较简单，直接上代码：

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

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