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`

**Explanation**:

**Explanation**:

"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`

**Explanation**:`"leeto" did not occur in "leetcode", so we return -1.`

**Explanation**:

"leeto" did not occur in "leetcode", so we return -1.

-`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)`.

classSolution:

defstrStr(self,haystack:str,needle:str) ->int:

for iinrange(len(haystack)):

j=0

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

j+=1

if j==len(needle):

return i

varstrStr=function (haystack,needle) {

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

let j=0

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

j+=1

if (j==needle.length) {

return i

}

j+=1

if (j==needle.length) {

return i

}

}

}

};