Programming Discussion Thread

[shuffle2]

http://stackoverflow.com/questions/4363665/hashmap-implementation-to-count-the-occurences-of-each-character

I came up with this: http://ideone.com/edaq3X
(Note that I haven't used Java for a long time, and I hate it's guts...this exercise reminded me just how much it sucks)

Code:

import java.util.*;
import java.lang.*;

class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        Map hitcounts = new HashMap();
        String uppercase_alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        
        for (char uppercase_char : uppercase_alphabet.toCharArray())
        {
            hitcounts.put(uppercase_char, 0);
        }
        
        String input_string = "Make This WhatEveR input You Want";
        
        for (char character : input_string.toCharArray())
        {
            if (hitcounts.containsKey(character))
            {
                hitcounts.put(character, (int)hitcounts.get(character) + 1);
            }
        }
        
        System.out.println(hitcounts);
    }
}

You don't need to initialize a hashmap, but in this case it doesn't really matter.

Edit: Here is a version that doesn't initialize the hashmap: http://ideone.com/k88VPa

Code:

import java.util.*;
import java.lang.*;

class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        Map hitcounts = new HashMap();
        String input_string = "Make This WhatEveR input You Want";
        
        for (char character : input_string.toCharArray())
        {
            if (Character.isUpperCase(character))
            {
                int hitcount = 0;
                if (hitcounts.containsKey(character))
                {
                    hitcount = (int)hitcounts.get(character);
                }
                
                hitcounts.put(character, hitcount + 1);
            }
        }
        
        System.out.println(hitcounts);
    }
}

Note that this time the hashmap will not contain keys with a value of 0. You would need to account for this when accessing the hashmap, or insert the keys explicitly.