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
没什么头绪的题，经点播发现可以找规律，如果repeat A超过B的长度还不能contain B，就足以说明组不成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;
}
}
