计算相同和重复单词的数量

Question

我的代码计算给定字符串上的关键字数量，但我需要将重复的关键字计为 1，并将其计为唯一关键字。有人可以帮我吗，我不知道该怎么做 :( .请问？

#include <stdio.h>
#include <string.h>

int main()
{
  char string[] = "\nauto else else do while >> \nfor 1+2 == goto";
  char *keywords[] =
  { "auto", "break", "else", "case", "if" };
  int s = 0, i, x = 1;
  char *result;
  char delims[] = "  \n";
  int keyword = 0, ukeyword = 0;

  printf("Input: %s", string);
  //Count the number of lines
  for (i = 0; string[i] != '\0'; ++i)
  {
    if (string[i] == '\n')
      ++s;
  }
  printf("\n\nNo. of lines: %d", s);

  result = strtok(string, delims);
  while (result != NULL )
  {
    for (x = 0; x != 5; x++)
    {
      if (strcmp(result, keywords[x]) == 0)
      {
        keyword++;
      }
    }

    result = strtok(NULL, delims);
  }

  printf("\nNo. of keywords: %d.\n", keyword);
  printf("\nNo. of unique keywords: %d.\n", ukeyword);
  return 0;
}

所以输出应该是:

Input:
auto else else do while >>
for 1+2 == goto

No. of lines:2
No. of keywords:3
No. of unique keywords:2

Answer 1

您必须分别计算每个关键字的出现次数:

int keyword_count[5];

并且当您找到关键字时，您必须增加相应的计数器:

keywords ++;
keyword_count[x] ++;

最后唯一关键字是那些具有 keyword_count[x]==1 的关键字