leetcode 686. Repeated String Match

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.

题目大意：
A需要重复至少几遍才能使得B是A的子串
思路：
当B的长度小于A的长度时，直接将A扩大两倍与B进行匹配
当B的长度大于A的长度时，直接将A扩大三倍与B进行匹配，获得匹配的位置i,那么结果就是(i + 1)/A.size()或者(i + 1)/A.size() + 1

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
        string k = "";
        if (B.size() <= A.size()) k = A + A;
        else {
            int x = B.size() / A.size() + 2;
            for (int i = 0; i < x; ++i) {
                k += A;
            }
        }
        int j = 0;
        for (int i = 0; i < k.size(); ++i) {
            if (k[i] == B[j]) {
                ++j;
                if (j == B.size()) {
                    if ((i+1) % A.size() == 0) return (i + 1)/A.size();
                    return (i + 1)/A.size() + 1;
                }
            } else {
                j = 0;
            }
        }
        return -1;
    }
};