[leetcode]686. Repeated String Match

[leetcode]686. Repeated String Match

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Analysis

微博第一次中奖~—— [嘻嘻，耶~]

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”).
主要用到了AA.find(B) != string::npos

Implement

方法一：（天知道这个while循环的条件我纠结了多久！！！）

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
        int res = 1;
        int lenA = A.length();
        int lenB = B.length();
        int len = 0;
        string AA = A;
        while(len <= lenB+lenA){
            if(AA.find(B) != string::npos)
                return res;
            AA += A;
            res++;
            len += lenA; 
        }
        return -1;
    }
};

方法二：

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
        int res = 1;
        int lenA = A.length();
        int lenB = B.length();
        string AA = A;
        while(AA.length() < lenB){ 
            AA += A;
            res++;
        }
        if(AA.find(B) != string::npos)
                return res;
        AA += A;
        res++;
        if(AA.find(B) != string::npos)
                return res;
        return -1;
    }
};