LeetCode 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.

题意：给定两个字符串A、B，如果B是A重复数次组成的字符串的子串，返回重复次数；如果不是，返回-1.

思路：

定义一个新字符串str，在str里重复增加字符串A，直到大于等于字符串B的长度。如果str包含子串B，返回true；否则，str再增加一个字符串A，再判断是否包含子串B。对str再加一个A来处理：A= “abcd”，B= “dabcdabcda”这种情况，即当str的长度大于等于B后，最多再加一个字符串就可以判断重复字符串是否可以包含B。代码如下：

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