字串尋找之KMP

根據定義next[0]=-1，假設next[j]=k, 即P[0...k-1]==P[j-k,j-1]

   1)若P[j]==P[k]，則next[j+1]=k+1;

   2)若P[j]!=P[k]，則k=next[k]。

#include <stdio.h>#include <string.h>#include <stdlib.h>void get_next(char *pat, int next[]){        int i, j, plen;        plen = strlen(pat);        i = 0;        j = -1;        next[0] = -1;        while(i < plen-1){                if(j == -1 || pat[i] == pat[j]){                        ++i;                        ++j;                        next[i] = j;                }                else{                        j = next[j];                }        }}int kmp_match(char *str, char *pat){        int slen, plen, i, j;        int *next;        if(str == NULL || pat == NULL){                return -1;        }        slen = strlen(str);        if(slen <= 0){                return -1;        }        plen = strlen(pat);        if(plen <= 0){                return -1;        }        if(slen < plen){                return 0;        }        next = (int *)malloc(plen*sizeof(int));        if(next == NULL){                return -1;        }        get_next(pat, next);        i = 0;        j = 0;        while(i < slen){                if(j == -1 || str[i] == pat[j]){                        i++;                        j++;                }                else{                        j = next[j];                }                if(j == plen){                        free(next);                        return (i-plen);                }        }        free(next);        return 0;}int main(int argc, char *argv[]){        int ret;        if(argc != 3){                printf("usage: a.out [patter] [string] \n");                return -1;        }        ret = kmp_match(argv[2], argv[1]);        printf("ret = %d\n", ret);        return 0;}