【leetcode】Implement strStr()

Question：

 

Implement strStr().

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

 

Anwser 1：  O(n*m）

 

class Solution {

public:

    char *strStr(char *haystack, char *needle) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int haylen = strlen(haystack);

        int needlen = strlen(needle);

        

        for(int i = 0; i <= haylen - needlen; i++){

            char *p = haystack + i;

            char *q = needle;

            while(*q != '\0'){

                if(*p != *q){

                    break;

                } else {

                    p++;

                    q++;

                }

            }

            

            if(*q == '\0'){

                return haystack + i;

            }

        }

        

        return NULL;

    }

};

 

Anwser 2：  O(n + m) KMP

 

class Solution {

public:

    char *strStr(char *haystack, char *needle) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int haylen = strlen(haystack);

        int needlen = strlen(needle);

        int* fail = new int[needlen];

        memset(fail, -1, needlen * sizeof(int));    // strlen(fail)

        

        int i, j, k;

        for (i = 1; i < needlen; ++i) {

            for (k = fail[i-1]; k >= 0 && needle[i] != needle[k+1]; k = fail[k]);

            if (needle[k+1] == needle[i])

                fail[i] = k + 1;

        }

        

        i = j = 0;

      

       while(i < haylen && j < needlen)      // while(haystack[i] && needle[j])

        {

            if (haystack[i] == needle[j])

            {

                ++i;

                ++j;

            }

            else if(j == 0) ++i;

            else j = fail[j-1] + 1;

        }

        

        delete fail;

        

        /*

        if (needle[j]) {

            return NULL;

        }  else {

            return haystack + i - j;

        }*/

        if(j == needlen){

            return haystack + i - j;

        } else {

            return NULL;

        }

    }

};

 

 

参考推荐：

KMP字符串匹配算法