686.md

Repeated String Match

Description

link

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Solution

• 只能是1次或者2次

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Code

O(n)

class Solution:
    def repeatedStringMatch(self, A: str, B: str) -> int:
        times = -(-len(B) // len(A))
        for i in range(2):
            if B in A*(times + i):
                return times + i
        return -1