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;
}
}
};
