使用C语言解决字符串匹配问题的方法
最常想到的方法是使用KMP字符串匹配算法:
#include <stdio.h> #include <stdlib.h> #include <string.h> int get_nextval(char *pattern, int next[]) { //get the next value of the pattern int i = 0, j = -1; next[0] = -1; int patlen = strlen(pattern); while ( i < patlen - 1){ if ( j == -1 || pattern[i] == pattern[j]){ ++i; ++j; if (pattern[i] != pattern[j]) next[i] = j; else next[i] = next[j]; } else j = next[j]; } return(0); } int kmpindex(char *target, char *pattern, int pos) { int tari = pos, pati = 0; int tarlen = strlen(target), patlen = strlen(pattern); int *next = (int *)malloc(patlen * sizeof(int)); get_nextval(pattern, next); while ( tari < tarlen && pati < patlen ){ if (pati == -1 ||target[tari] == pattern[pati]){ ++tari; ++pati; }else{ pati = next[pati]; } } if(next != NULL) free(next); next = NULL; if (pati == patlen) return tari - pati; else return -1; } int main() { char target[50], pattern[50]; printf("imput the target:\n" ); scanf("%s",target); printf("imput the pattern:\n" ); scanf("%s",pattern); int ans = kmpindex(target,pattern,0); if (ans == -1) printf("error\n"); else printf("index:%d\n",ans); return 0; }
练习题
题目描述:
读入数据string[ ],然后读入一个短字符串。要求查找string[ ]中和短字符串的所有匹配,输出行号、匹配字符串。匹配时不区分大小写,并且可以有一个用中括号表示的模式匹配。如“aa[123]bb”,就是说aa1bb、aa2bb、aa3bb都算匹配。
输入:
输入有多组数据。
每组数据第一行输入n(1<=n<=1000),从第二行开始输入n个字符串(不含空格),接下来输入一个匹配字符串。
输出:
输出匹配到的字符串的行号和该字符串(匹配时不区分大小写)。
样例输入:
4
Aab
a2B
ab
ABB
a[a2b]b
样例输出:
1 Aab
2 a2B
4 ABB
ac代码
#include <stdio.h> #include <stdlib.h> #include <string.h> #define MAX 1001 #define LEN 101 struct str { char name[101]; }; int main() { struct str strs[MAX]; struct str t[LEN]; int i, n, len, j, k, left, right, count, flag; char text[LEN], newtext[LEN]; while (scanf("%d", &n) != EOF) { // 接收数据 getchar(); for (i = 0; i < n; i ++) { scanf("%s", strs[i].name); } // 接收文本串 getchar(); gets(text); len = strlen(text); for (i = left = right = 0; i < len; i ++) { if (text[i] == '[') { left = i; } else if (text[i] == ']') { right = i; break; } } count = right - left - 1; if (count <= 0) { // 没有正则匹配 for (i = j = 0; i < len; i ++) { if (text[i] != '[' && text[i] != ']') { newtext[j ++] = text[i]; } } newtext[j] = '\0'; for (i = 0; i < n; i ++) { if (strcasecmp(strs[i].name, newtext) == 0) { printf("%d %s\n", i + 1, strs[i].name); } } }else { // 需要正则匹配 for (j = 1, k = 0; j <= count; j ++, k ++) { // 构建文本数组 memset(t[k].name, '\0', sizeof(t[k].name)); for (i = 0; i < left; i ++) { t[k].name[i] = text[i]; } t[k].name[i] = text[left + j]; strcat(t[k].name, text + right + 1); } // 正则匹配 for (i = 0; i < n; i ++) { for (j = flag = 0; j < count; j ++) { if (strcasecmp(strs[i].name, t[j].name) == 0) { flag = 1; break; } } if (flag) { printf("%d %s\n", i + 1, strs[i].name); } } } } return 0; }
/**************************************************************
Problem: 1165
User: wangzhengyi
Language: C
Result: Accepted
Time:0 ms
Memory:948 kb
****************************************************************/
赞 (0)