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"

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

輸入字串s和正整數k，對於每2k個字母，將其中前k個字母翻轉。最後如果剩餘字母少於k個，則將其全部翻轉；若多於k個，則將前k個翻轉，其餘不變。

思路：利用void reverse(str.begin(),str.end())函式直接原位翻轉，若要翻轉到其他變數，可使用str2.assign(str.rbegin(), str.rend())函式。

    string reverseStr(string s, int k) {
        int left = s.size()%(2*k);
        int i;
        
        if (left<k)
        {
            for(i=0;i<s.size()-left;i+=(2*k))
                reverse(s.begin()+i,s.begin()+i+k);
            reverse(s.begin()+i,s.end());
        }
        else
        {
            for(i=0;i<s.size()-left;i+=(2*k))
                reverse(s.begin()+i,s.begin()+i+k);
            reverse(s.begin()+i,s.begin()+i+k);
        }
        return s;
    }