# LeetCode 28. Implement strStr()



Implement [strStr()](http://www.cplusplus.com/reference/cstring/strstr/).

Return the index of the first occurrence of needle in haystack, or `-1` if `needle` is not part of `haystack`.

**Clarification:**

What should we return when `needle` is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when `needle` is an empty string. This is consistent to C's [strstr()](http://www.cplusplus.com/reference/cstring/strstr/) and Java's [indexOf()](https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf\(java.lang.String\)).

**Example 1:**

```
Input: haystack = "hello", needle = "ll"
Output: 2
```

**Example 2:**

```
Input: haystack = "aaaaa", needle = "bba"
Output: -1
```

**Example 3:**

```
Input: haystack = "", needle = ""
Output: 0
```

**Constraints:**

* `0 <= haystack.length, needle.length <= 5 * 10^4`
* `haystack` and `needle` consist of only lower-case English characters.

## Solution:

[English Version in Youtube](https://youtu.be/G9c18kcusBY)

[中文版解答Youtube Link](https://youtu.be/2XJeC0eVNrc)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1CA411K7f8/)

```
class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.length(), n = needle.length();
        for (int i = 0; i < m - n + 1; i++) {
            int j = 0;
            for (; j < n; j++) {
                if (haystack[i + j] != needle[j]) {
                    break;
                }
            }
            if (j == n) {
                return i;
            }
        }
        return -1;
    }
};
```

