Find the most popular words in a file - Java version

wordcount.java

import java.io.*;
import java.util.*;

public class wordcount
{
	public static class Word implements Comparable<Word>
	{
		String word;
		int count;

		@Override
		public int hashCode()
		{
			return word.hashCode();
		}

		@Override
		public boolean equals(Object obj)
		{
			return word.equals(((Word)obj).word);
		}

		@Override
		public int compareTo(Word b)
		{
			return b.count - count;
		}
	}

	public static void main(String[] args)
	throws IOException
	{
		long time = System.currentTimeMillis();

		Map<String, Word> countMap = new HashMap<String, Word>();

		BufferedReader reader = new BufferedReader(new InputStreamReader(new FileInputStream(args[0])));
		String line;
		while ((line = reader.readLine()) != null) {
			String[] words = line.split("[^A-ZÅÄÖa-zåäö]+");
			for (String word : words) {
				if ("".equals(word)) {
					continue;
				}

				Word wordObj = countMap.get(word);
				if (wordObj == null) {
					wordObj = new Word();
					wordObj.word = word;
					wordObj.count = 0;
					countMap.put(word, wordObj);
				}

				wordObj.count++;
			}
		}

		reader.close();

		SortedSet<Word> sortedWords = new TreeSet<Word>(countMap.values());
		int i = 0;
		for (Word word : sortedWords) {
			if (i > 10) {
				break;
			}

			System.out.println(word.count + "\t" + word.word);

			i++;
		}

		time = System.currentTimeMillis() - time;

		System.out.println("in " + time + " ms");
	}
}

Very nice !

It is incorrect if you have multiple words with the same occurence. SortedMap uses compareTo and if you only use the count in that method, you will lose those words that have the same count...