如何实现字谜和回文功能来检查用户输入的单词？

Question

I got some help earlier fixing up one of the functions I am using in this program, but now I'm at a loss of logic.

I have three purposes and two functions in this program. The first purpose is to print a sentence that the user inputs backwards. The second purpose is to check if any of the words are anagrams with another in the sentence. The third purpose is to check if any one word is a palindrome.

I successfully completed the first purpose. I can print sentences backwards. But now I am unsure of how I should implement my functions to check whether or not any words are anagrams or palindromes.

Here's the code;

/*
 * Ch8pp14.c
 *
 *  Created on: Oct 12, 2013
 *      Author: RivalDog
 *      Purpose: Reverse a sentence, check for anagrams and palindromes
 */

#include <stdio.h>
#include <ctype.h> //Included ctype for tolower / toupper functions
#define bool int
#define true 1
#define false 0

//Write boolean function that will check if a word is an anagram
bool check_anagram(char a[], char b[])
{
   int first[26] = {0}, second[26] = {0}, c = 0;
// Convert arrays into all lower case letters
   while(a[c])
   {
       a[c] = (tolower(a[c]));
       c++;
   }

   c = 0;

   while(b[c])
      {
       b[c] = (tolower(b[c]));
       c++;
      }

      c = 0;

   while (a[c] != 0)
   {
      first[a[c]-'a']++;
      c++;
   }

   c = 0;

   while (b[c] != 0)
   {
      second[b[c]-'a']++;
      c++;
   }

   for (c = 0; c < 26; c++)
   {
      if (first[c] != second[c])
         return false;
   }

   return true;
}

//Write boolean function that will check if a word is a palindrome
bool palindrome(char a[])
{
    int c=0, j, k;
    //Convert array into all lower case letters
    while (a[c])
    {
        a[c] = (tolower(a[c]));
        c++;
    }

    c = 0;
    j = 0;
    k = strlen(a) - 1;
    while (j < k)
    {
        if(a[j++] != a[k--])
            return false;
    }

    return true;
}

int main(void)
{
    int i = 0, j = 0, k = 0;
    char a[80], terminator;
    //Prompt user to enter sentence, store it into an array
    printf("Enter a sentence: ");
    j = getchar();
    while (i < 80)
    {
        a[i] = j;
        ++i;
        j = getchar();
        if (j == '!' || j == '.' || j == '?')
        {
            terminator = j;
            break;
        }
        else if(j == '\n')
        {
            break;
        }
    }
    while(a[k])
    {
        a[k] = (tolower(a[k]));
        k++;
    }
    k = 0;
    while(k < i)
    {
        printf("%c", a[k]);
        k++;
    }
    printf("%c\n", terminator);
    //Search backwards through the loop for the start of the last word
    //print the word, and then repeat that process for the rest of the words
    for(j = i; j >= 0; j--)
    {
        while(j > -1)
        {
            if (j == 0)
            {
                for(k=j;k<i;k++)
                    {
                        printf("%c", a[k]);
                    }
                printf("%c", terminator);
                    break;
            }
            else if (a[j] != ' ')
                --j;
            else if (a[j] == ' ')
                {
                    for(k=j+1;k<i;k++)
                        {
                            printf("%c", a[k]);
                        }
                    printf(" ");
                        break;
                }
        }
        i = j;
    }
    //Check if the words are anagrams using previously written function
    for( i = 0; i < 80; i++)
    {
        if (a[i] == ' ')
        {

        }
    }

    //Check if the words are palindromes using previously written function

return 0;
}

I was thinking that perhaps I could again search through the array for the words by checking if the element is a space, and if it is, store from where the search started to the space's index-1 in a new array, repeat that process for the entire sentence, and then call my functions on all of the arrays. The issue I am seeing is that I can't really predict how many words a user will input in a sentence... So how can I set up my code to where I can check for anagrams/palindromes?

Thank you everyone!

~RivalDog

Answer 1

Would be better,if you first optimize your code and make it readable by adding comments.Then you can divide the problem in smaller parts like 1.How to count words in a string? 2.How to check whether two words are anagrams? 3.How to check whether a word is palindrome or not? And these smaller programs you could easily get by Googling. Then your job will be just to integrate these answers. Hope this helps.

Answer 2

To check anagram, no need to calculate number of words and comparing them one by one or whatever you are thinking.
Look at this code. In this code function read_word() is reading word/phrase input using an int array of 26 elements to keep track of how many times each letter has been seen instead of storing the letters itself. Another function equal_array() is to check whether both array a and b (in main ) are equal (anagram) or not and return a Boolean value as a result.

#include <stdio.h>
#include <ctype.h>
#include <stdbool.h>

void read_word(int counts[26]);
bool equal_array(int counts1[26],int counts2[26]);

int main()
{
    int a[26] = {0}, b[26] = {0};

    printf("Enter first word/phrase: ");
    read_word(a);
    printf("Enter second word/phrase: ");
    read_word(b);

    bool flag = equal_array(a,b);
    printf("The words/phrase are ");
    if(flag)
        printf("anagrams");
    else
        printf("not anagrams"); 

    return 0;
}

void read_word(int counts[26])
{
    int ch;
    while((ch = getchar()) != '\n')
    if(ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z')
        counts[toupper(ch) - 'A']++;
}

bool equal_array(int counts1[26],int counts2[26])
{
    int i = 0;
    while(i < 26)
    {
        if(counts1[i] == counts2[i])
            i++;
        else
            break;  
    }

    return i == 26 ? true : false;  
}