Repeated String Match【重复字符串匹配】

PROBLEM：

Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.

For example, with A = "abcd" and B = "cdabcdab".

Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").

Note:
The length of A and B will be between 1 and 10000.

SOLVE：

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
        for (auto i = 0, j = 0; i < A.size(); ++i) {
            for (j = 0; j < B.size() && A[(i + j) % A.size()] == B[j]; ++j){};
            if (j == B.size()) 
                return (i + j) / A.size() + ((i + j) % A.size() != 0 ? 1 : 0);
        }
        return -1;
    }
};

思路：从A中不同字符开始，与B字符串比较，超出尾部时循环；当顺利比较遍历完B字符串则输出结果；否则当A字符串遍历完一次时输出-1.