[LeetCode] 686. Repeated String Match 重复字符串匹配

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.

给2个字符串，找到字符串A需要重复的次数，使得字符串B是字符串A的子串，如果没有答案，则返回-1。

解法1:  Brute fore. a modified version of string find, which does not stop at the end of A, but continue matching by looping through A

解法2: KMP, O(n + m) version that uses a prefix table (KMP)

解法3: Rabin-Karp Algorithm

Java: 1

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class Solution {     public int repeatedStringMatch(String A, String B) {         StringBuilder sb = new StringBuilder();         sb.append(A);         int count = 1;         while(sb.indexOf(B)<0){             if(sb.length()-A.length()>B.length()){                 return -1;             }             sb.append(A);             count++;         }                   return count; }

Python: 1

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class Solution(object):     def repeatedStringMatch(self, A, B):         """         :type A: str         :type B: str         :rtype: int         """         sa, sb = len(A), len(B)         x = 1         while (x - 1) * sa <= 2 * max(sa, sb):             if B in A * x: return x             x += 1         return -1

Python: 3

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# Time:  O(n + m) # Space: O(1)   class Solution(object):     def repeatedStringMatch(self, A, B):         """         :type A: str         :type B: str         :rtype: int         """         def check(index):             return all(A[(i+index) % len(A)] == c                        for i, c in enumerate(B))           M, p = 10**9+7, 113         p_inv = pow(p, M-2, M)         q = (len(B)+len(A)-1) // len(A)           b_hash, power = 0, 1         for c in B:             b_hash += power * ord(c)             b_hash %= M             power = (power*p) % M           a_hash, power = 0, 1         for i in xrange(len(B)):             a_hash += power * ord(A[i%len(A)])             a_hash %= M             power = (power*p) % M           if a_hash == b_hash and check(0): return q           power = (power*p_inv) % M         for i in xrange(len(B), (q+1)*len(A)):             a_hash = (a_hash-ord(A[(i-len(B))%len(A)])) * p_inv             a_hash += power * ord(A[i%len(A)])             a_hash %= M             if a_hash == b_hash and check(i-len(B)+1):                 return q if i < q*len(A) else q+1           return -1

C++: 1

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int repeatedStringMatch(string A, string B) {     for (auto i = 0, j = 0; i < A.size(); ++i) {         for (j = 0; j < B.size() && A[(i + j) % A.size()] == B[j]; ++j);         if (j == B.size()) return (i + j) / A.size() + ((i + j) % A.size() != 0 ? 1 : 0);     }     return -1; }

C++: 2

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int repeatedStringMatch(string a, string b) {     vector<int> prefTable(b.size() + 1); // 1-based to avoid extra checks.     for (auto sp = 1, pp = 0; sp < b.size(); ) {       if (b[pp] == b[sp]) prefTable[++sp] = ++pp;       else if (pp == 0) prefTable[++sp] = pp;       else pp = prefTable[pp];     }           for (auto i = 0, j = 0; i < a.size(); i += max(1, j - prefTable[j]), j = prefTable[j]) {         while (j < b.size() && a[(i + j) % a.size()] == b[j]) ++j;         if (j == b.size()) return (i + j) / a.size() + ((i + j) % a.size() != 0 ? 1 : 0);     }     return -1; }

C++:

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class Solution { public:     int repeatedStringMatch(string A, string B) {         string t = A;         for (int i = 1; i <= B.size() / A.size() + 2; ++i) {             if (t.find(B) != string::npos) return i;             t += A;         }         return -1;     } };

C++:

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class Solution { public:     int repeatedStringMatch(string A, string B) {         int m = A.size(), n = B.size();         for (int i = 0; i < m; ++i) {             int j = 0;             while (j < n && A[(i + j) % m] == B[j]) ++j;             if (j == n) return (i + j - 1) / m + 1;         }         return -1;     } };

　　

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