LeetCode #459 Repeated Substring Pattern 重复的子字符串

Description:
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:
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.)

题目描述:
给定一个非空的字符串，判断它是否可以由它的一个子串重复多次构成。给定的字符串只含有小写英文字母，并且长度不超过10000。

示例:
示例 1:

输入: "abab"

输出: True

解释: 可由子字符串 "ab" 重复两次构成。
示例 2:

输入: "aba"

输出: False
示例 3:

输入: "abcabcabcabc"

输出: True

解释: 可由子字符串 "abc" 重复四次构成。 (或者子字符串 "abcabc" 重复两次构成。)

思路:

1. 类似找素数的思路, 从 1个字符开始匹配到 1/2个长度的字符
   时间复杂度O(n ^ 2), 空间复杂度O(n)
2. 考虑到 s是重复串, s + s也是重复串, 破坏 s + s的头尾两个字符, 剩下的如果包含 s, 则说明 s中有重复的子字符串
   e.g.

• s = "aba" s + s = "abaaba" -> "baab" -> false
• s = "abab" s + s = "abababab" -> "babababa" -> true
  时间复杂度O(n), 空间复杂度O(n)

3. 正则表达式, 将要匹配的子字符串当作一个组, 匹配到结尾, 这个组需要出现次数大于 1
   时间复杂度O(n ^ 2), 空间复杂度O(n)

代码:
C++:

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        return (s + s).substr(1, s.size() * 2 - 2).find(s) != -1;
    }
};

Java:

class Solution {
    public boolean repeatedSubstringPattern(String s) {
        return (s + s).substring(1, s.length() * 2 - 1).indexOf(s) != -1;
    }
}

Python:

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        return re.match(r"([a-z]+)\1+$", s) != None