關鍵字出現個數

問題

編寫 一個程序，統計輸入中C語言每個關鍵字的出現次數。

算法分析

While ( 仍有新單詞讀入）
    If(新單詞是否是關鍵字）
        相應關鍵字出現次數加1；
輸出關鍵字及出現次數；

定義一個結構用以表示關鍵字與其出現次數：

struct   Key {
    char *keyword;
    int count;
};

關鍵字表的組織：使用一個有序(提高效率)的結構數組來存放關鍵字表及關鍵字出現次數：

struct   Key   Keytab[ ] = {
  "auto", 0,
  "break",0,
  "case", 0,
  ...
  "while", 0
};

在有序數據集中查找指定數據項最常用及最快的算法是折半查找算法

算法實現

#include <stdio.h>
struct Key {
  char  *keyword;
  int  keycount;
} Keytab[ ] = {
  "auto", 0,
  "break", 0,
  "case", 0,
  ...
  "while", 0
};
#define  MAXWORD  20
#define  NKEYS  (sizeof(Keytab) / sizeof(struct  Key))
#define  LETTER  'a'
#define  DIGIT  '0'

struct  Key  *binary(char  *word,  struct Key  tab[ ],  int  n);
char  getword(char  *w,  int lim);
char  type( int c);
void printKey(struct Key  tab[ ],  int  n);

int main( )	/* count  C keyword */
{
  int t;
  char word[MAXWORD];
  struct Key *p;
 
  while((t = getword(word, MAXWORD)) != EOF)
  if( t = = LETTER)
    if(( p = binary(word, Keytab, NKEYS)) != NULL)
      p->keycount++;
  printKey(keytab, NKEYS);
  return 0;
}

struct Key *binary(char *word, struct Key tab[ ], int n)
{
  int cond;
  struct Key *low = &tab[0];
  struct Key *high = &tab[n-1];
  struct Key *mid;
 
  while(low <= high){
    mid = low + (high – low) / 2;
    if((cond = strcmp(word, mid->keyword)) < 0)
       high = mid – 1;
    else if ( cond > 0)
       low = mid + 1;
    else
       return (mid);
  }
  return (NULL);
}

char getword(char *w, int lim)
{
  int c, t;
 
  if(type(c = *w++ = getchar( )) != LETTER){
    *w = '\0';
    return ( c);
  }
  while(--lim > 0) {
    t = type(c = *w++ = getchar( ));
    if( t != LETTER && t != DIGIT){
      ungetc(c,stdin);
      break;
    }
  }
  *(w-1) = '\0';
  return ( LETTER);
}

char type(int c)	/* return type of ASCII character */
{
  if( c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z')
    return ( LETTER );
  else if ( c >= '0' && c <= '9')
    return ( DIGIT );
  else return (c);
}	

void printKey(struct Key  tab[ ],  int  n)
{
  struct Key *p;
  for(p=Keytab, p < Keytab+n; p++)
    if(p->keycount > 0)
       printf("%4d%s\n", p->keycount, p->keyword);
}