Java
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/** 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. */ public class Repeated_String_Match { class Solution { public int repeatedStringMatch(String A, String B) { int count = 1; StringBuilder Acopy = new StringBuilder(A); for (; Acopy.length() < B.length(); ) { Acopy.append(A); count++; } if (Acopy.indexOf(B) >= 0) { return count; } if (Acopy.append(A).indexOf(B) >= 0) { return count+1; } return -1; } } }
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// OJ: https://leetcode.com/problems/repeated-string-match/ // Time: O(B * (A + B)) // Space: O(A + B) // Ref: https://leetcode.com/problems/repeated-string-match/solution/ class Solution { public: int repeatedStringMatch(string a, string b) { int k = 1; string s = a; for (; s.size() < b.size(); ++k) s += a; if (s.find(b) != string::npos) return k; return (s + a).find(b) != string::npos ? k + 1 : -1; } };
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class Solution(object): def repeatedStringMatch(self, A, B): """ :type A: str :type B: str :rtype: int """ na, nb = len(A), len(B) times = nb / na + 3 for i in range(1, times): if B in A * i: return i return -1
Java
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class Solution { public int repeatedStringMatch(String A, String B) { int lengthA = A.length(), lengthB = B.length(); StringBuffer sb = new StringBuffer(); int maxLength = lengthA * 2 + lengthB; int repeatTimes = 0; while (sb.length() <= maxLength) { sb.append(A); repeatTimes++; if (sb.toString().indexOf(B) >= 0) return repeatTimes; } return -1; } }
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// OJ: https://leetcode.com/problems/repeated-string-match/ // Time: O(B * (A + B)) // Space: O(A + B) // Ref: https://leetcode.com/problems/repeated-string-match/solution/ class Solution { public: int repeatedStringMatch(string a, string b) { int k = 1; string s = a; for (; s.size() < b.size(); ++k) s += a; if (s.find(b) != string::npos) return k; return (s + a).find(b) != string::npos ? k + 1 : -1; } };
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class Solution(object): def repeatedStringMatch(self, A, B): """ :type A: str :type B: str :rtype: int """ na, nb = len(A), len(B) times = nb / na + 3 for i in range(1, times): if B in A * i: return i return -1