Implement strStr()

Given a haystack string and a needle string, find the first place (starting from 0) where the needle string appears in the haystack string. If not, returns - 1.

Example 1:

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

Example 2:

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

Explain:

When need is an empty string, what value should we return? This is a good question in an interview.

For this problem, we should return 0 when need is an empty string. This matches the C language's strstr() and Java's indexOf() definitions.

Solution 1: use built-in functions

class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.empty()) return 0;
        int pos=haystack.find(needle);//Find needle in the string haystack
        if(pos==-1) return -1;//find() not found returned - 1
        return pos;
    }
};

Solution 2: use KMP algorithm

class Solution {
public:
    vector<int> next(string str)
	    {
	        vector<int> v;
	        v.push_back(-1);//Insert an element at the end of the vector.
	        int i = 0, j = -1;
	        while (i < str.size())
	        {
	           if (j == -1 || str[i] == str[j])
	      	{
	      	 ++i;
	       	 ++j;
	          v.push_back(j);
	        } 
	        else
	        j = v[j];
	        }
	          return v;
	        }
	    int strStr(string haystack, string needle) 
	    {
	        int i = 0；//Main string location
	        int j = 0；//Mode string position
	        int len1 = haystack.size(), len2 = needle.size();
	        vector<int> nextptr
	        if(needle.empty())
	            return 0;
	        
	        nextptr = next(needle); 
	        
	        while((i < len1)&&(j < len2))
	        {
	            
	            if((j == -1) || (haystack[i] == needle[j]))
	            //When j=-1, move the position of the main string, and skip the equivalent characters
	            {
	                i++;
	                j++;
	            }
	            else
	            {
	                j = nextptr[j];//Get next matching location
	            }	      
	       }
	          
	        if (j == needle.size())
	            return i - j;
	        else 
	            return -1;
	    }
};