168.Repeated String Match（重复的字符串匹配）

题目：

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.

给定两个字符串A和B，找到A必须重复的最小次数，使得B是它的子字符串。 如果没有这样的解决方案，返回-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").

返回3，因为重复A三次（“abcdabcdabcd”），B是它的子串; 和B不是A重复两次的子串（“abcdabcd”）。

Note:
The length of A and B will be between 1 and 10000.

A和B的长度在1到10000之间。

解答：

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

详解：