459. Repeated Substring Pattern(easy)

最新推荐文章于 2020-07-25 01:42:52 发布

SourDumplings

最新推荐文章于 2020-07-25 01:42:52 发布

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本文介绍了一种通过正则表达式和步进法检测字符串是否由重复子串构成的方法。利用正则表达式的match方法及步进法遍历检查，实现了高效地判断字符串是否由某一子串重复拼接而成。

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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: changzheng300@foxmail.com
 * @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: changzheng300@foxmail.com
 * @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;
    }
};