具有霍夫曼树的优先级队列

Question

i am trying to create a Huffman tree by reading in a file and counting the frequency of each letter space symbol etc. i'm using a Priorityqueue to queue the items from smallest to largest but when i insert them into the queue they dont queue correctly here is my code. package huffman;

import java.io.FileNotFoundException; import java.io.FileReader; import java.util.ArrayList; import java.util.PriorityQueue; import java.util.Scanner;

public class Huffman {

public ArrayList<Frequency> fileReader(String file)
{
    ArrayList<Frequency> al = new ArrayList<Frequency>();
    Scanner s;
    try {

        s = new Scanner(new FileReader(file)).useDelimiter("");
        while (s.hasNext())
        {
            boolean found = false;
            int i = 0;
            String temp = s.next();
            while(!found)
            {


                if(al.size() == i && !found)
                {
                    found = true;
                    al.add(new Frequency(temp, 1));
                }
                else if(temp.equals(al.get(i).getString()))
                {
                    int tempNum = al.get(i).getFreq() + 1;
                    al.get(i).setFreq(tempNum);
                    found = true;
                }
                i++;

            }



        }
    } catch (FileNotFoundException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return al;
}
public void buildTree(ArrayList<Frequency> al)
{
    PriorityQueue<Frequency> pq = new PriorityQueue<Frequency>();
    for(int i = 0; i < al.size(); i++)
    {
        pq.add(al.get(i));          
    }
    while(pq.size() > 0)
    {
        System.out.println(pq.remove().getString());
    }
}
public void printFreq(ArrayList<Frequency> al)
{
    for(int i = 0; i < al.size(); i++)
    {
        System.out.println(al.get(i).getString() + "; " + al.get(i).getFreq());
    }
}

}

in the buildTree() method is where im having the problem. what im trying to do is queue Frequency objects which holds the letter/space/symbol and the frequency as an int the frequency class is this. public class Frequency implements Comparable { private String s; private int n;

Frequency(String s, int n)
{
    this.s = s;
    this.n = n;
}
public String getString()
{
    return s;
}
public int getFreq()
{
    return n;
}
public void setFreq(int n)
{
    this.n = n;
}
@Override
public int compareTo(Object arg0) {
    // TODO Auto-generated method stub
    return 0;
}

}

how can i get the priorityqueue to use the frequency number to queue them from smallest to biggest?

Answer 1

Actually you missed to implement the compareTo method to make your object effectively comparable.

The compareTo method, as documentation states, should

return a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

This means that in your case you should do something like:

public int compareTo(Object arg0)
{
  Frequency other = (Frequency)arg0;

  return n < other.n ? -1 : (n == other.n ? 0 : 1);
}

But mind that comparable has a generic type that is preferable: Comparable<T> so you can avoid the cast on arg0 to make it a Frequency object with static type safety too:

class Frequency implements Comparable<Frequency> {   
  public int compareTo(Frequency f2) {
    // directly compare
  }
}

Answer 2

I think that "Auto-generated method stub" needs to be filled in with a real implementation of a "compareTo" so as to satisfy the requirements for something to be Comparable, which I assume the PriorityQueue is going to rely upon. The implementation is probably going to be "n < arg0", with appropriate downcasting from Object.

Answer 3

A Priority Queue , just as a data structure, is based on the concept of an ordering - you use such a structure when you want to order elements in a certain way - which elements are more important than others, etc.

In Java, ordering objects is usually done in one of two ways - your objects implement the Comparable interface, or you supply a Comparator<E> which knows how to order objects of type E .

To determine which object is "more important" than another, the compareTo() method is invoked. This method has a pretty simple contract:

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

Your implementation of Frequency.compareTo() always returns 0 for the comparison. Thus, you are specifying that all Frequency objects are equal to any other Frequency objects. This is clearly not what you want.