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.

Code (Java):

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class Solution {     public int repeatedStringMatch(String A, String B) {         if (A == null || A.length() == 0) {             return -1;         }           if (B == null || B.length() == 0) {             return 0;         }           // step 1: calculate the min number of steps since lenA >= lenB         //         int count = 0;         StringBuilder sb = new StringBuilder();           while (sb.length() <= B.length() + A.length()) {             if (sb.toString().contains(B)) {                 break;             } else {                 sb.append(A);                 count++;             }         }           if (sb.toString().contains(B)) {             return count;         } else {             return -1;         }     } }

Discussion:
Why we need to check sb().length <= A.length() + B.length()? There is one corner case
A = abc
B = cabcacbc

The expected answer is 4. 

If we just stop when sb.length() > B.length(), we may miss one more step.