686. Repeated String Match --------M

Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b​​​​​​ to be a substring of a after repeating it, return -1.

Notice: string "abc" repeated 0 times is "",  repeated 1 time is "abc" and repeated 2 times is "abcabc".

 

Example 1:

Input: a = "abcd", b = "cdabcdab"
Output: 3
Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it.

Example 2:

Input: a = "a", b = "aa"
Output: 2

Example 3:

Input: a = "a", b = "a"
Output: 1

Example 4:

Input: a = "abc", b = "wxyz"
Output: -1

 

Constraints:

• 1 <= a.length <= 104
• 1 <= b.length <= 104
• a and b consist of lower-case English letters.

A:

就是重复，然后再查看。最多做string match 2次

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class Solution { public: int repeatedStringMatch(string A, string B) { int nA = A.length(), nB = B.length(); string ss = ""; int res = 0; while(ss.length()<= nB + nA){ ss += A; res++; if(ss.length() >= nB && (ss.find(B) != string::npos)) return res; } return -1; } };