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459-repeated-substring-pattern.md Added Chinese solutions.
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‎.gitignore‎

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‎en/1-1000/459-repeated-substring-pattern.md‎

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# 459. Repeated Substring Pattern - LeetCode Solution
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LeetCode English link: [459. Repeated Substring Pattern](https://leetcode.com/problems/repeated-substring-pattern)
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# 459. Repeated Substring Pattern - LeetCode Solution Best Practice
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LeetCode link: [459. Repeated Substring Pattern](https://leetcode.com/problems/repeated-substring-pattern), difficulty: **Easy**.
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LeetCode Chinese link: [459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern)
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[中文题解](#中文题解)
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## LeetCode problem description
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## Description of "459. Repeated Substring Pattern"
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Given a string `s`, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.
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Difficulty: **Easy**
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### [Example 1]
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**Input**: `s = "abcabcabcabc"`
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- `s` consists of lowercase English letters.
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## Intuition behind the Solution
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[中文题解](#中文题解)
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The key to solving this problem is to see clearly that if `s` can be obtained by repeating the substring, then the starting letter of the substring must be `s[0]`.
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Once you understand this, the scope of substring investigation is greatly narrowed.
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## 力扣“459. 重复的子字符串”问题描述
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力扣链接：[459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern), 难度: **简单**。
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给定一个非空的字符串 `s` ，检查是否可以通过由它的一个子串重复多次构成。
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### [示例 1]
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**输入**: `s = "abcabcabcabc"`
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**输出**: `true`
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**解释**: `可由子串 "abc" 重复四次构成。 (或子串 "abcabc" 重复两次构成。)`
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### [示例 2]
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**输入**: `s = "aba"`
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**输出**: `false`
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# 中文题解
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## 思路
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解决本问题的关键是要看清楚一点：通过子串的重复能得到`s`，那么子串的起始字母一定是`s[0]`。想明白了这一点，子串的排查范围就大大缩小了。

‎zh/1-1000/459-repeated-substring-pattern.md‎

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# 459. Repeated Substring Pattern - LeetCode Solution
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LeetCode English link: [459. Repeated Substring Pattern](https://leetcode.com/problems/repeated-substring-pattern)
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LeetCode Chinese link: [459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern)
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[中文题解](#中文题解)
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## LeetCode problem description
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Given a string `s`, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.
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# 459. 重复的子字符串 - 力扣题解最佳实践
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力扣链接：[459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern), 难度: **简单**。
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Difficulty: **Easy**
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## 力扣“459. 重复的子字符串”问题描述
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给定一个非空的字符串 `s` ，检查是否可以通过由它的一个子串重复多次构成。
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### [Example 1]
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**Input**: `s = "abcabcabcabc"`
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### [示例 1]
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**输入**: `s = "abcabcabcabc"`
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**Output**: `true`
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**输出**: `true`
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**Explanation**: `It is the substring "abc" four times or the substring "abcabc" twice.`
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**解释**: `可由子串 "abc" 重复四次构成。 (或子串 "abcabc" 重复两次构成。)`
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### [Example 2]
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**Input**: `s = "aba"`
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### [示例 2]
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**输入**: `s = "aba"`
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**Output**: `false`
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**输出**: `false`
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### [Constraints]
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### [约束]
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- `1 <= s.length <= 10000`
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- `s` consists of lowercase English letters.
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## Intuition behind the Solution
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[中文题解](#中文题解)
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- `s` 由小写英文字母组成
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The key to solving this problem is to see clearly that if `s` can be obtained by repeating the substring, then the starting letter of the substring must be `s[0]`.
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Once you understand this, the scope of substring investigation is greatly narrowed.
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## 思路
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解决本问题的关键是要看清楚一点：通过子串的重复能得到`s`，那么子串的起始字母一定是`s[0]`。想明白了这一点，子串的排查范围就大大缩小了。
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## Complexity
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* Time: `O(N * N)`.
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* Space: `O(N)`.
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## 复杂度
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* 时间：`O(N * N)`。
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* 空间：`O(N)`。
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## Python
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```python
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## 力扣“459. 重复的子字符串”问题描述
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力扣链接：[459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern), 难度: **简单**。
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给定一个非空的字符串 `s` ，检查是否可以通过由它的一个子串重复多次构成。
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### [示例 1]
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**输入**: `s = "abcabcabcabc"`
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**输出**: `true`
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**解释**: `可由子串 "abc" 重复四次构成。 (或子串 "abcabc" 重复两次构成。)`
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### [示例 2]
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**输入**: `s = "aba"`
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**输出**: `false`
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# 中文题解
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## 思路
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解决本问题的关键是要看清楚一点：通过子串的重复能得到`s`，那么子串的起始字母一定是`s[0]`。想明白了这一点，子串的排查范围就大大缩小了。

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