Leetcode | Repeated String Match
2017-12-10 20:11
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原题链接:https://leetcode.com/problems/repeated-string-match
题意内容如下所示:
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.
题目大意就是给定两个字符串A和B,若有i个A字符串相连构成的字符串,假定为s,能够使得B成为s的子字符串,求i的最小值,若不能成为s的子字符串,则返回-1。
题意非常的清晰明了,就是用最少次数的A字符串构成一个新的字符串s,使得B能够成为s的子串。判断一个字符串是否是另一个的子串,在C++中可以直接用string头文件中已经定义好的find函数,而C语言就可以用经典的KMP算法。所以我们要解决的主要问题就是A最多需要重复几次才能让B成为其子串。
假设存在A的某个子串为a,所以通过B的长度/A的长度,剩下的就是A的一些子串,所以字符串B就存在两个模式Aa或aA,aAa,所以在决定最多重复次数时就定为B的长度/A的长度+2。
源代码如下所示:
题意内容如下所示:
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.
题目大意就是给定两个字符串A和B,若有i个A字符串相连构成的字符串,假定为s,能够使得B成为s的子字符串,求i的最小值,若不能成为s的子字符串,则返回-1。
题意非常的清晰明了,就是用最少次数的A字符串构成一个新的字符串s,使得B能够成为s的子串。判断一个字符串是否是另一个的子串,在C++中可以直接用string头文件中已经定义好的find函数,而C语言就可以用经典的KMP算法。所以我们要解决的主要问题就是A最多需要重复几次才能让B成为其子串。
假设存在A的某个子串为a,所以通过B的长度/A的长度,剩下的就是A的一些子串,所以字符串B就存在两个模式Aa或aA,aAa,所以在决定最多重复次数时就定为B的长度/A的长度+2。
源代码如下所示:
class Solution { public: int repeatedStringMatch(string A, string B) { string s = A; for (int i = 1; i <= B.length()/A.length()+2; i++, s+=A) if (s.find(B) != string::npos) return i; < 8c33 span class="hljs-keyword">return -1; } };
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