leetcode 686. Repeated String Match

题目：

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.

http://blog.youkuaiyun.com/zsheng_/article/details/78153614

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

！！！这里补充find函数知识点：

1.find()：查找第一次出现的目标字符串，如果查找成功则输出查找到的第一个位置，否则返回-1；

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1. #include<iostream>  
2. #include<csdtio>  
3. using namespace std;  
4.   
5. int main(){  
6. string s1 = "abcdef";  
7. string s2 = "de";  
8. int ans = s1.find(s2) ;   //在S1中查找子串S2  
9. cout<<ans<<endl;  
10. }  

查找从指定位置开始的第一次出现的目标字符串：

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1. #include<iostream>  
2. #include<csdtio>  
3. using namespace std;  
4.   
5. int main(){  
6. string s1 = "abcdef";  
7. string s2 = "de";  
8. int ans = s1.find(s2, 2) ;   //从S1的第二个字符开始查找子串S2  
9. cout<<ans<<endl;  
10. }