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.

Solution

沒什么頭緒的題，經點播發(fā)現可以找規(guī)律，如果repeat A超過B的長度還不能contain B，就足以說明組不成B了。

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;
        if(sb.append(A).toString().contains(B)) return ++count;
        return -1;
    }
}