[LeetCode] 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:

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

对字符串进行反转，规则是：每隔k个反转k个字母。如果剩下的字母不满足k则，则将剩下的字母全部反转。取t = n / k，将字符串分成k份，每隔一份反转一次。

class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.size(), t = n / k;
        for (int i = 0; i <= t; i++) {
            if (i % 2 == 0) {
                if (i * k + k < n)
                    reverse(s.begin() + i * k, s.begin() + i * k + k);
                else
                    reverse(s.begin() + i * k, s.end());
            }
        }
        return s;
    }
};
// 6 ms

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