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

Example 1:

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

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

分析：

返回haystack中首次出現needle的索引，不存在則返回-1，needle為空則返回0。依次比較j與i中字元，若完全相同，則返回相應的下標，否則繼續與i中下一字元比較，直至遍歷完i中所有字元。

class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.size() == 0)
            return 0;
        if(haystack.size() < needle.size())
            return -1;
        int i = 0;//haystack下標
        int j = 0;//needle下標
        while(i <= haystack.size()-1 && j <= needle.size()-1) 
        {
            if(haystack[i] == needle[j])
            {
                i++;
                j++;
            }
            else
            {
                i = i-j+1;
                j = 0;
            }
        
        }
        if(j == needle.size())
            return i-j;
        else
            return -1;
        
    }
};