Original link: [leetcoder.net - LeetCoder: Fucking Good LeetCode Solutions](https://leetcoder.net/en/leetcode/28-find-the-index-of-the-first-occurrence-in-a-string)

LeetCode 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), Difficulty: **Easy**.

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`.

**Input**: `haystack = "sadbutsad", needle = "sad"`

**Output**: `0`

"sad" occurs at index 0 and 6.

The first occurrence is at index 0, so we return 0.

**Input**: `haystack = "leetcode", needle = "leeto"`

**Output**: `-1`

- `1 <= haystack.length, needle.length <= 10000`

- `haystack` and `needle` consist of only lowercase English characters.

- This kind of question can be solved with one line of code using the built-in `index()`. Obviously, the questioner wants to test our ability to control the loop.

- For `heystack`, traverse each character in turn. There may be two situations:

- This question is easier to understand by looking at the code directly.

- Time complexity: `O(N + M)`.

- Space complexity: `O(1)`.

class Solution:

def strStr(self, haystack: str, needle: str) -> int:

for i in range(len(haystack)):

j = 0

while i + j < len(haystack) and haystack[i + j] == needle[j]:

j += 1

if j == len(needle):

return i

return -1

var strStr = function (haystack, needle) {

for (let i = 0; i < haystack.length; i++) {

let j = 0

while (i + j < haystack.length && haystack[i + j] == needle[j]) {

}

}

return -1

};