leetcode 686. Repeated String Match

最新推荐文章于 2022-07-26 11:05:48 发布

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最新推荐文章于 2022-07-26 11:05:48 发布

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本文链接：https://blog.youkuaiyun.com/icurious/article/details/78325505

本文探讨了如何寻找字符串A最少重复次数，使得字符串B成为其子串的问题。通过实例讲解了算法实现细节，并给出了具体代码示例。

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.

来源： https://leetcode.com/problems/repeated-string-match/description/

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
       // A = "abcd" and B = "cdabcdab". return 3

        if (A==B)return 1;
        int i=1;
        int la=A.length();
        int lb=B.length();

        if (la>(lb<<1)){
            if (A.find(B)!=string::npos)
                return 1;
        }
        else{//la<=lb
            string backup=A;
            for (i = 1; i <=(lb/la)+2 ; backup+=A,++i) {
                //cout<<backup<<" i="<<i<<endl;
                int newLen=backup.length();
                if (newLen<lb)continue;
                else{//newLen>=lb
                    if (backup.find(B)!=string::npos){
                        return i;
//                        break;
                    }//backup.find(B)==string::npos
                }
            }
        }

        return -1;

    }
};