leetcode 686. Repeated String Match

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;    }};

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;    }};