leetcode 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.

直接使用C++的find函数做查找即可，当A的长度小于B的时候，我们可以先进行重复A，直到A的长度大于等于B，并且累计次数cnt。那么此时我们用find来找，看B是否存在A中，如果存在直接返回cnt。如果不存在，我们再加上一个A，再来找，这样可以处理这种情况A=”abc”, B=”cab”，如果此时还找不到，说明无法匹配，返回-1，

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution {public:    int repeatedStringMatch(string a, string b)    {        int count = 1;        string aa = a;        while (aa.length() < b.length())        {            count++;            aa += a;        }        if (aa.find(b) != aa.npos)            return count;        else        {            aa += a;            if (aa.find(b) != aa.npos)                return count + 1;            else                return -1;        }    }};