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.

 

重复几次A后，B是A的子串

 

C++（16ms）：

 1 class Solution {
 2 public:
 3     int repeatedStringMatch(string A, string B) {
 4         string t = A ;
 5         for(int i = 1 ; i <= B.length()/A.length()+2 ; i++){
 6             if (t.find(B) != string::npos){
 7                 return i ;
 8             }else{
 9                 t += A ;
10             }
11         }
12         return -1; 
13     }
14 };

 

Java(412ms):

 1 class Solution {
 2     public int repeatedStringMatch(String A, String B) {
 3         String t = A ;
 4         for(int i = 1 ; i <= B.length()/A.length()+2 ; i++){
 5             if (t.indexOf(B) != -1)
 6                 return i ;
 7             else
 8                 t += A ;
 9         }
10         return -1 ;
11     }
12 }

 

posted @ 2017-10-17 19:35  __Meng  阅读(208)  评论(0)  编辑  收藏  举报