686. Repeated String Match

Description

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.
Problem URL

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Solution

给一个String A， B，判断B是否是A重复多次的子字符串，返回重复的次数。

Using stringbuilder build a string which is longer than B. Add A to find whether it has B inside. If not, return -1.

Code

class Solution {
    public int repeatedStringMatch(String A, String B) {
        StringBuilder sb = new StringBuilder();
        int count = 0;
        while (sb.length() < B.length()){
            sb.append(A);
            count++;
        }
        if (sb.toString().contains(B)){
            return count;
        }
        if (sb.append(A).toString().contains(B)){
            return ++count;
        }
        return -1;
    }
}

Time Complexity: O(n)
Space Complexity: O(n)

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