Anagram Tester

[Zelaine]

Hello!
This program down below lets you enter two strings, then compares them and outputs if they are anagrams of each other or not. However, I think that it's a little bit too long and wonder if it can be done in an easier and/or shorter way.

(And no, this is not my homework or anything like that, I'm just doing this all alone of sheer curiosity because I want to learn Java, believe me or not.)

Thanks in advance! :)

import java.util.*;
import javax.swing.*;

public class tjena{
	public static void main(String[] args){
		String t1 = JOptionPane.showInputDialog("Enter a string of text"), s1;
		s1 = t1.toUpperCase(); s1 = s1.replaceAll("\\s", "");
		
		String t2 = JOptionPane.showInputDialog("Enter another string of text"), s2;
		s2 = t2.toUpperCase(); s2 = s2.replaceAll("\\s", "");
		
		boolean anagram = anagramTest(s1, s2);
		t1 = anagram ? "Your strings are anagrams of each other!" : "Your strings are not anagrams of each other!";
		JOptionPane.showMessageDialog(null, t1);
	}
	
	static boolean anagramTest(String s1, String s2){
		boolean anagram = false;
		if(s1.length() == s2.length()){
			
			int firstAnagram[] = alphabetTester(s1), secondAnagram[] = alphabetTester(s2);

			for(int x=0;x<firstAnagram.length;x++){
				if(firstAnagram[x] != secondAnagram[x])
					break;
				else if(firstAnagram[x] == secondAnagram[x] && x == firstAnagram.length - 1)
					anagram = true;
			}
			
		}
		return anagram;
	}
	
	static int[] alphabetTester(String s){ // THIS PART RETURNS AN ARRAY WITH THE LENGTH OF THE ALPHABET! THE CELLS CONTAIN
		int test[] = new int[25];		   // THE AMOUNT OF A:S, B:S, C:S, ETC. THAT THE STRING CONTAINED IN ALPHABETICAL ORDER
		for(int x=65, z = 0;x<=90;x++,z++){
			for(int y=0;y<s.length();y++){
				if(s.charAt(y) == (char)x)
					test[z]++;
			}
		}
		return test;
	}
}

[Norm]

One trick for counting characters whose values are contiguous is to use the values of the characters to compute the index. If 'a' is the first char and 'z' is the last, then the index values would be: 'a'=0, 'b'=1, 'c'=2 etc
Compute these indexes by: theChar - 'a' where theChar is a char in the range 'a' to 'z'

BTW Hardcoding 65 and 90 makes the code hard to understand. It is better to code 'A' and 'Z'

[jim829]

Hello! ...wonder if it can be done in an easier and/or shorter way.

Yes, there is a much simpler way. I came up with this many years ago but I'm certain someone else has done it too. All anagrams of each other sort to the same string. Just convert the strings into separate character arrays and sort each array. Then just iterate thru the arrays comparing the characters in each position. If they are all the same, the two strings are anagrams of each other. This can be done after the length checks.

Regards,
Jim

[Zelaine]

Is this what you meant Jim?

import java.util.*;

public class tjena{
	public static void main(String[] args){
		System.out.print("Enter a string of words: ");
		char word1[] = transformingString();
		System.out.print("Enter another string of words: ");
		char word2[] = transformingString();
		word1 = sorter(word1); word2 = sorter(word2);
		String s = Arrays.equals(word1, word2) ? "Your strings are anagrams of each other!" : "Your strings are not anagrams of each other!";
		System.out.print(s);
	}
	
	static char[] sorter(char word[]){
		for(int x=0, z=0;x<=25;x++)
			for(int y=0;y<word.length;y++)
				if((char)(x+65) == word[y]){
					char c = word[y];
					word[y] = word[z];
					word[z] = c;
					z++;
				}
		return word;
	}
	
	static char[] transformingString(){
		Scanner hej  = new Scanner(System.in);
		String s = hej.nextLine();
		s = s.replaceAll("\\s+", ""); String s1 = s.toUpperCase();
		char test[] = s1.toCharArray();
		return test;
	}
}