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-[459. Repeated Substring Pattern](solutions/1-1000/459-repeated-substring-pattern.md)

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LeetCode English link:[28. Find the Index of the First Occurrence in a String](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string),

LeetCode English link:[28. Find the Index of the First Occurrence in a String](https://leetcode.com/problems/find-the-index-of-the-first-occurrence-in-a-string)

LeetCode Chinese link:[28. 找出字符串中第一个匹配项的下标](https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string)

([中文题解](#中文题解))

[中文题解](#中文题解)

Given two strings`needle` and`haystack`, return the**index** of the first occurrence of`needle` in`haystack`, or`-1` if`needle` is not part of`haystack`.

LeetCode English link:[459. Repeated Substring Pattern](https://leetcode.com/problems/repeated-substring-pattern)

LeetCode Chinese link:[中文问题名称](https://leetcode.cn/problems/repeated-substring-pattern)

[中文题解](#中文题解)

Given a string`s`, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

Difficulty:**Easy**

**Input**:`s = "abcabcabcabc"`

**Output**:`true`

**Explanation**:`It is the substring "abc" four times or the substring "abcabc" twice.`

**Input**:`s = "aba"`

**Output**:`false`

-`s` consists of lowercase English letters.

[中文题解](#中文题解)

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]`.

Once you understand this, the scope of substring investigation is greatly narrowed.

* Time:`O(N * N)`.

* Space:`O(N)`.

classSolution:

defrepeatedSubstringPattern(self,s:str) ->bool:

for iinrange(1,int(len(s)/2)+1):

iflen(s)% i==0and s[:i]*int(len(s)/ i)== s:

varrepeatedSubstringPattern=function (s) {

for (let i=1; i<=s.length/2; i++) {

if (s.length% i!=0) {

}

if (s.slice(0, i).repeat(s.length/ i)== s) {

}

}

};

力扣链接：[459. 重复的子字符串](https://leetcode.cn/problems/repeated-substring-pattern), 难度:**简单**。

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。

**输入**:`s = "abcabcabcabc"`

**输出**:`true`

**解释**:`可由子串 "abc" 重复四次构成。 (或子串 "abcabc" 重复两次构成。)`

**输入**:`s = "aba"`

**输出**:`false`

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。