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

Traverse the string once, and if the`needle.length` characters after the current position are equal to`needle`, return the current`index`.

- 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:

1. First, the character is not equal to the first letter of`needle`. Then process the next character.

2. Second, if the character is equal to the first letter of`needle`, continue to compare the next character of`heystack` and`needle` in an internal loop until they are not equal or`needle` has completely matched.

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

* Time:`O(m* n)`.

* Time:`O(m+ n)`.

* Space:`O(n)`.

classSolution:

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

for iinrange(len(haystack)):

if haystack[i:i+len(needle)]== needle:

return i

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++) {

if (haystack.slice(i, i+needle.length)== needle) {

return i

let j=0

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

j+=1

if (j==needle.length) {

return i

}

}

}

-`haystack` 和`needle` 仅由小写英文字符组成

遍历一遍字符串，如果从当前位置起的后的`needle.length`个字符等于`needle`，则返回当前的`index`。

- 这样的题目，如果用内置的`index()`，一行代码就可以实现。显然，出题人是想考察我们对循环的控制能力。

- 针对`heystack`，依次遍历每个字符。可能出现两种情况：

- 本题直接看答案比较容易理解。

* 时间：`O(m * n)`。