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686. Repeated String Match

时间:2018-10-19 18:07:17      阅读:134      评论:0      收藏:0      [点我收藏+]

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;
    }

 

686. Repeated String Match

原文:https://www.cnblogs.com/jessie2009/p/9817922.html

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