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.

class Solution:
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        if len(set(A)) < len(set(B)):
            return -1
        for i in range(1,len(B)+1):
            t = A*i
            if t.find(B)!=-1:
                return i
        return -1

686. Repeated String Match

原文：https://www.cnblogs.com/bernieloveslife/p/10245822.html