Question

I am trying to find the three highest values in a TreeMap. I wrote a code that is kind of doing it, but I would like to ask whether you can suggest a more efficient way.
Basically, I am saving each word of my text in a TreeMap along with the number of times it appears in the text. Then I am using a comparator to sort the values. Then I am iterating through the newly created Map until I reach the last three values, which are the highest values after the sorting and print them out. I am going to use large texts, so this is not a very good way.
Here is my code:

class Text{
    public static void main(String args[]) throws FileNotFoundException, IOException{
        final File textFile = new File("C://FileIO//cinderella.txt"); 
        final BufferedReader in = new BufferedReader(new FileReader(textFile));                               
        final TreeMap<String, Integer> frequencyMap = new TreeMap<String, Integer>(); 

        String currentLine; 
        while ((currentLine = in.readLine()) != null) {  
            currentLine = currentLine.toLowerCase();  
            final StringTokenizer parser = new StringTokenizer(currentLine, " \t\n\r\f.,;:!?'"); 
            while (parser.hasMoreTokens()) { 
                final String currentWord = parser.nextToken(); 
                Integer frequency = frequencyMap.get(currentWord); 
                if (frequency == null) { 
                    frequency = 0; 
                } 
                frequencyMap.put(currentWord, frequency + 1);
            } 
        }  

        System.out.println("This the unsorted Map: "+frequencyMap);

        Map sortedMap = sortByComparator(frequencyMap);
        int i = 0;
        int max=sortedMap.size();
        StringBuilder query= new StringBuilder();

        for (Iterator it = sortedMap.entrySet().iterator(); it.hasNext();) {
            Map.Entry<String,Integer> entry = (Map.Entry<String,Integer>) it.next();
            i++;
            if(i<=max && i>=(max-2)){
                String key = entry.getKey();
                //System.out.println(key);
                query.append(key);
                query.append("+");
            }
        }
        System.out.println(query);
    }

    private static Map sortByComparator(TreeMap unsortMap) {
        List list = new LinkedList(unsortMap.entrySet());

        //sort list based on comparator
        Collections.sort(list, new Comparator() {
            public int compare(Object o1, Object o2) {
                return ((Comparable) ((Map.Entry) (o1)).getValue())
                       .compareTo(((Map.Entry) (o2)).getValue());
            }
        });

        //put sorted list into map again
        Map sortedMap = new LinkedHashMap();
        for (Iterator it = list.iterator(); it.hasNext();) {
            Map.Entry entry = (Map.Entry)it.next();
            sortedMap.put(entry.getKey(), entry.getValue());

        }
        return  sortedMap;
    }   
}

Answer 1

I would count the frequencies with a hash map, and then loop over them all, selecting the top 3. You minimize comparisons this way, and never have to sort. Use the Selection Algorithm

-edit, the wikipedia page details many different implementations of the selection algorithm. To be specific, just use a bounded priority queue, and set the size to 3. Dont get fancy and implement the queue as a heap or anything. just use an array.