题目：

Implement strStr().

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

思路：

BF或者KMP算法。BF无法通过时间复杂度要求。

代码：

class Solution {
public:
    int* kmp;
    char *strStr(char *haystack, char *needle) {
        if(haystack==NULL || needle == NULL)
        {
            return NULL;
        }
        else if(haystack==needle)
        {
            return haystack;
        }
        else
        {
            calKMP(needle);
            int len1=getLen(haystack);
            int len2=getLen(needle);
            int i=0;
            int j=0;
            while(j<len2&&i<len1)
            {
                if(*(haystack+i)==*(needle+j))
                {
                    i++;
                    j++;
                }
                else
                {
                    
                    if(j==0)
                    {
                        i++;
                    }
                    else
                    {
                        j=kmp[j-1]+1;
                    }
                }
            }
            if(j==len2)
            {
                return haystack+i-len2;
            }
            else
            {
                return NULL;
            }
        }
    }
    
    void calKMP(char *needle) {
        int len = getLen(needle);
        kmp= new int[len];
        kmp[0]=-1;
        for(int i=1;i<len;i++)
        {
            int cur=i-1;
            kmp[i]=-1;
            while(cur!=-1)
            {
                if(*(needle+kmp[cur]+1)==*(needle+i))
                {
                    kmp[i]=kmp[cur]+1;
                    break;
                }
                else
                {
                    cur=kmp[cur];
                }
            }
        }
    }
    
    int getLen(char *needle) {
        if(needle==NULL)
        {
            return 0;
        }
        else
        {
            int len=0;
            while(*needle!='\0')
            {
                len++;
                needle++;
            };
            return len;
        }
    }
};