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.

思路：想要B包含在A的倍数n字符串中，n必须等于len(B)//len(A)或者len(B)//len(A)+1

代码：python3

import math
class Solution:
    def repeatedStringMatch(self, A: str, B: str) -> int:
        lA,lB=len(A),len(B)
        count = math.ceil(lB/lA)
        s=count*A
        print(B)
        print(s)
        return count*(B in s) or (count+1)*(B in (count+1)*A) or -1