[leetcode-686-Repeated String Match]

标签：amp   turn   note   minimum   npos   string   and   length   get   

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]

标签：amp   turn   note   minimum   npos   string   and   length   get   

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