The problem is:

Implement strStr().

Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.

A naive method: (1) find the the first letter of needle in haystack , (2)comparing other consecutive characters of needle and haystack, (3) if all letters in needle can be found in haystack consecutively, then return the position in haystack . 

class Solution {
public:
    char *strStr(char *haystack, char *needle) {
		if(haystack==NULL || needle==NULL)
			return NULL;

		if(needle[0]=='\0')
			return haystack;

		// Find the first character of needle in haystack
		int hay_len=strlen(haystack);
		int need_len=strlen(needle);
		int pos=0;
		bool found=false;
		while(hay_len-pos>=need_len && haystack[pos]!='\0') {
			if(haystack[pos] == needle[0]){
				int th=pos+1;
				int tn=1;
				while(haystack[th] == needle[tn]) {
					if(needle[tn]=='\0')
						break;
					th++;
					tn++;
				}
				if(tn == need_len) {
					found=true;
					break;
				}
			}
			pos++;
		}

		if(found)
			return haystack+pos;
		else
			return NULL;
    }
};

A better solution is to use the famous KMP algorithm( http://www.geeksforgeeks.org/searching-for-patterns-set-2-kmp-algorithm/ ). 

class Solution {
public:
    char *strStr(char *haystack, char *needle) {
		if(haystack==NULL || needle==NULL)
	 		return NULL;
	 	if(needle[0]=='\0')
	 		return haystack;

		int n=strlen(haystack);
		int m=strlen(needle);
		vector<int> pi(m); // the longest proper prefix which is also suffix
		computePrefix(needle,pi);
		//printPrefix(pi); // TEST ONLY

		int q=0; // index for needle
		int i=0; // index for haystack
		while(i<n) {
			//cout<<"needle["<<q<<"]="<<needle[q]<<", haystack["<<i<<"]="<<haystack[i]<<endl; // TEST ONLY
			if(needle[q]==haystack[i]) {
				q++;
				i++;
			}
			if(q==m) {
				//cout<<"Pattern occurs with shift "<<i-m<<endl; // TEST ONLY
				return haystack+i-m;
			} else if(needle[q]!=haystack[i]) {
				if(q!=0)
					q=pi[q-1];
				else
					i++;
			}
		}
		return NULL;
	}

	// Calculate the longest proper prefix which is also suffix.
	// Examples:
	// 	For the pattern “AABAACAABAA”, lps[] is [0, 1, 0, 1, 2, 0, 1, 2, 3, 4, 5]
	//	For the pattern “ABCDE”, lps[] is [0, 0, 0, 0, 0]
	//	For the pattern “AAAAA”, lps[] is [0, 1, 2, 3, 4]
	//	For the pattern “AAABAAA”, lps[] is [0, 1, 2, 0, 1, 2, 3]
	//	For the pattern “AAACAAAAAC”, lps[] is [0, 1, 2, 0, 1, 2, 3, 3, 3, 4]
	void computePrefix(char* P, vector<int>& pi) {
		int m=strlen(P);
		pi[0]=0; // lps[0] is always 0
		int k=0; // length of the previous prefix suffix
		int i=1;
		while(i<m) {
			if(P[k]==P[i]) {
				k++;
				pi[i]=k;
				i++;
			} else {
				if(k!=0) {
					k=pi[k-1];
				} else {
					pi[i]=0;
					i++;
				}
			}
		}
	}

	void printPrefix(vector<int>& pi) {
		cout<<"pi={ ";
		for(auto& x:pi)
			cout<<x<<" ";
		cout<<"}"<<endl;
	}
};