541. 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 = 2Output: "bacdfeg"Restrictions:
- The string consists of lower English letters only.
- Length of the given string and k will in the range [1, 10000]
这道题是说一个字符串s按照每隔2k个字符（从0开始计算）的offset开始旋转k个字符。如果剩余的字符没有达到k的长度就全部旋转，如果超过k的长度但是没有达到2k的长度，就只旋转k个长度的字符。
这里设置i为起始旋转点，j为旋转子串的循环下标。right为每个旋转子串的最右边节点下标。这里注意到每个头尾节点满足下标加起来等于i + right。程序如下：
class Solution {public: string reverseStr(string s, int k) { string res = s; if (s.size() == 1) return s; for (int i = 0; i < s.size(); i += k * 2) { int right = min(i + k - 1, (int)s.size() - 1); for (int j = i; j <= (i + right) / 2; ++j) { char t = res[j]; res[j] = res[i + right - j]; res[i + right - j] = t; } } return res; }};