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Commit1595fb3

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

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.idea/
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temp/
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en/empty-template.md
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zh/empty-template.md
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en/empty-template-en.md
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zh/empty-template-zh.md

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

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#459. Repeated Substring Pattern - LeetCode Solution
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LeetCodeEnglishlink:[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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