[leetcode] 686. Repeated String Match @ python

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原题

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.

解法

B有可能比较长, 我们尝试构造一个C, 长度是len(B)//len(A) + 2个A, 然后检查B是否在C里, 如果在, 尝试减少ans, 查看B是否在C里.

代码

class Solution(object):
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        ans = len(B)//len(A) + 2
        C = A*ans
        if B not in C:
            return -1
        while B in A*(ans-1):
            ans -= 1
        return ans