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[leetcode-686-Repeated String Match]

时间:2017-10-07 19:11:00      阅读:257      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.

思路:

最多重复次数为 

 b.length() / a.length() + 2;
int repeatedStringMatch(string a, string b) 
{
    string as = a;
    for (int rep = 1; rep <= b.length() / a.length() + 2; rep++, as += a)
        if (as.find(b) != string::npos) return rep;
    return -1;      
 }  

参考:

https://discuss.leetcode.com/topic/105565/c-java-clean-code

[leetcode-686-Repeated String Match]

原文:http://www.cnblogs.com/hellowooorld/p/7635395.html

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