[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

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

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

# 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

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

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++:

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++:

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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