459. Repeated Substring Pattern(easy)

Easy

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Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.

 

Example 1:

Input: "abab"

Output: True

Explanation: It's the substring "ab" twice.

Example 2:

Input: "aba"

Output: False

Example 3:

Input: "abcabcabcabc"

Output: True

Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)

 

使用正则表达式匹配:

Java:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-26 16:16:54
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/repeated-substring-pattern/
 * 
 * 直接使用String的match方法利用正则表达式进行匹配每一个前缀子串，
 * 注意先判断前缀子串的长度是否是总长度的约数进行剪枝，否则会超时
 */

class Solution
{
    public boolean repeatedSubstringPattern(String s)
    {
        int l = s.length();
        for (int i = 1; i <= l / 2; i++)
        {
            if (l % i == 0 && s.matches("(" + s.substring(0, i) + ")+"))
            {
                return true;
            }
        }
        return false;
    }
}

 尝试步进法:

C++:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-26 16:07:37
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/repeated-substring-pattern/
 * 
 * 思路就是尝试第一个子串和第二个之间的距离，看能否找到一个，时间复杂度为O(n^2)
 */

class Solution
{
public:
    bool repeatedSubstringPattern(string s)
    {
        int l = s.length();
        for (int d = 1; d < l; d++)
        {
            for (int i = 0; i < l - d; ++i)
            {
                if (s[i] != s[i + d])
                {
                    break;
                }
                if (l % d == 0 && i + d == l - 1)
                {
                    return true;
                }
            }
        }
        return false;
    }
};