Thread: anagram program help

Originally Posted by m37h0d

that one works todd, except that anagrams don't count things like whitespace and punctuation

anagrams are word plays - they don't ever count anything!

Oh, you mean the program. Yes, well, unfortunately, it counts everything. Therefore, use what you want and ignore the rest.

edit: i got it to work using this code .. now does anyone know if i should just count the letter .. or can an anagram use numbers as well?

Code:

#include <iostream>
#include <cctype>
#include <cstring>
#include <stdio.h>
#include <string.h> 
#include <stdlib.h> 

using namespace std;


struct characters { 
	int letters[256] ; 
} ; 

void map(const char * s, struct characters * c) { 
	memset( c, 0, sizeof(struct characters) ) ; 	
	while(*s) { 
		if (*s != ' ') { 
			c->letters[(int) tolower(*s)]++ ; 
		}
		s++ ; 
	} 
} 

int compare( const struct characters * c1, const struct characters * c2  ) { 
	return !memcmp( c1, c2, sizeof(struct characters)) ; 
} 


int main() {
	struct characters c1 , c2 ; 

	char s1[500];
	char s2[500];
	char mystring1[500];
	char mystring2[500];

	cout << "enter first string" << endl;
	cin.getline(mystring1, 500, '\n');
 int a=0;
	for (int i=0; mystring1[i] !='\0'; i++){
		if (isalpha(mystring1[i])){

			s1[a]=mystring1[i];
			a++;
		}
	}
s1[a]='\0';
	cout << s1 << endl;



	cout << "enter second string" << endl;
	cin.getline(mystring2, 500, '\n');

	int b=0;
	for (int i=0; mystring2[i] !='\0'; i++){
		if (isalpha(mystring2[i])){

			s2[b]=mystring2[i];
			b++;
		}
	}
s2[b]='\0';
	cout << s2 << endl;


	
	map(s1, &c1) ; 
	map(s2, &c2) ; 
	printf("The strings &#37;s Anagrams\n", (compare( &c1, &c2 ) ? "are" : "are not" ) ) ; 

	return 0 ; 




}

Well, since you need to be writing in C++, the only thing you should be taking out of my sample is the concept of how to do it.

Todd