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
    ****************************************************************/