标签：pen str space back for null col pac builder
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.
//Time: O(n), Space: O(1) public int repeatedStringMatch(String A, String B) { if (A == null || B == null) { return 0; } int count = 1; StringBuilder sb = new StringBuilder(A); 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 + 1; } return -1; }
标签：pen str space back for null col pac builder
原文地址：https://www.cnblogs.com/jessie2009/p/9817922.html
