A good Sorted List for Java

Question

I'm looking for a good sorted list for java. Googling around give me some hints about using TreeSet/TreeMap. But these components is lack of one thing: random access to an element in the set. For example, I want to access nth element in the sorted set, but with TreeSet, I must iterate over other n-1 elements before I can get there. It would be a waste since I would have upto several thousands elements in my Set.

Basically, I'm looking for some thing similar to a sorted list in .NET, with ability to add element fast, remove element fast, and have random access to any element in the list.

Has this kind of sorted list implemented somewhere? Thanks.

Edited

My interest in SortedList grows out of this problems: I need to maintains a list of many thousands object (and can grow up to many hundred of thousands). These objects will be persisted to database. I want to randomly select few dozens of element from the whole list. So, I tried to maintain a separated on-memory list that contains the primary keys (Long numbers) of all objects. I need to add/remove keys from the list when object is added / removed from database. I'm using an ArrayList right now but I'm afraid ArrayList would not suit it when the number of records grows. (Imagine you have to iterate over several hundred thousands of elements every time an object is removed from database). Back to the time when I did .NET programming, then I would use a sorted List (List is a .NET class that once Sorted property set to true, will maintain order of its element, and provide binary search that help remove/insert element very quick). I'm hoping that I can find some thing similar from java BCL but unluckily, I didn't find a good match.

Answer 1

Phuong:

Sorting 40,000 random numbers:

0.022 seconds

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Random;


public class test
{
    public static void main(String[] args)
    {
        List<Integer> nums = new ArrayList<Integer>();
        Random rand = new Random();
        for( int i = 0; i < 40000; i++ )
        {
            nums.add( rand.nextInt(Integer.MAX_VALUE) );
        }

        long start = System.nanoTime();
        Collections.sort(nums);
        long end = System.nanoTime();

        System.out.println((end-start)/1e9);
    }
}   

Since you rarely need sorting, as per your problem statement, this is probably more efficient than it needs to be.

Answer 2

To test the efficiancy of earlier awnser by Konrad Holl, I did a quick comparison with what I thought would be the slow way of doing it:

package util.collections;

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.ListIterator;

/**
 *
 * @author Earl Bosch
 * @param <E> Comparable Element
 *
 */
public class SortedList<E extends Comparable> implements List<E> {

    /**
     * The list of elements
     */
    private final List<E> list = new ArrayList();

    public E first() {
        return list.get(0);
    }

    public E last() {
        return list.get(list.size() - 1);
    }

    public E mid() {
        return list.get(list.size() >>> 1);
    }

    @Override
    public void clear() {
        list.clear();
    }

    @Override
    public boolean add(E e) {
        list.add(e);
        Collections.sort(list);
        return true;
    }

    @Override
    public int size() {
        return list.size();
    }

    @Override
    public boolean isEmpty() {
        return list.isEmpty();
    }

    @Override
    public boolean contains(Object obj) {
        return list.contains((E) obj);
    }

    @Override
    public Iterator<E> iterator() {
        return list.iterator();
    }

    @Override
    public Object[] toArray() {
        return list.toArray();
    }

    @Override
    public <T> T[] toArray(T[] arg0) {
        return list.toArray(arg0);
    }

    @Override
    public boolean remove(Object obj) {
        return list.remove((E) obj);
    }

    @Override
    public boolean containsAll(Collection<?> c) {
        return list.containsAll(c);
    }

    @Override
    public boolean addAll(Collection<? extends E> c) {

        list.addAll(c);
        Collections.sort(list);
        return true;
    }

    @Override
    public boolean addAll(int index, Collection<? extends E> c) {
        throw new UnsupportedOperationException("Not supported.");
    }

    @Override
    public boolean removeAll(Collection<?> c) {
        return list.removeAll(c);
    }

    @Override
    public boolean retainAll(Collection<?> c) {
        return list.retainAll(c);
    }

    @Override
    public E get(int index) {
        return list.get(index);
    }

    @Override
    public E set(int index, E element) {
        throw new UnsupportedOperationException("Not supported.");
    }

    @Override
    public void add(int index, E element) {
        throw new UnsupportedOperationException("Not supported.");
    }

    @Override
    public E remove(int index) {
        return list.remove(index);
    }

    @Override
    public int indexOf(Object obj) {
        return list.indexOf((E) obj);
    }

    @Override
    public int lastIndexOf(Object obj) {
        return list.lastIndexOf((E) obj);
    }

    @Override
    public ListIterator<E> listIterator() {
        return list.listIterator();
    }

    @Override
    public ListIterator<E> listIterator(int index) {
        return list.listIterator(index);
    }

    @Override
    public List<E> subList(int fromIndex, int toIndex) {
        throw new UnsupportedOperationException("Not supported.");
    }

}

Turns out its about twice as fast! I think its because of SortedLinkList slow get - which make's it not a good choice for a list.

Compared times for same random list:

• SortedLinkList : 15731.460
• SortedList : 6895.494
• ca.odell.glazedlists.SortedList : 712.460
• org.apache.commons.collections4.TreeList : 3226.546

Seems glazedlists.SortedList is really fast...

Answer 3

Depending on how you're using the list, it may be worth it to use a TreeSet and then use the toArray() method at the end. I had a case where I needed a sorted list, and I found that the TreeSet + toArray() was much faster than adding to an array and merge sorting at the end.

Answer 4

SortedList decorator from Java Happy Libraries can be used to decorate TreeList from Apache Collections. That would produce a new list which performance is compareable to TreeSet. https://sourceforge.net/p/happy-guys/wiki/Sorted%20List/

Answer 5

What about using a HashMap? Insertion, deletion, and retrieval are all O(1) operations. If you wanted to sort everything, you could grab a List of the values in the Map and run them through an O(n log n) sorting algorithm.

edit

A quick search has found LinkedHashMap, which maintains insertion order of your keys. It's not an exact solution, but it's pretty close.

Answer 6

Generally you can't have constant time look up and log time deletions/insertions, but if you're happy with log time look ups then you can use a SortedList.

Not sure if you'll trust my coding but I recently wrote a SortedList implementation in Java, which you can download from http://www.scottlogic.co.uk/2010/12/sorted_lists_in_java/. This implementation allows you to look up the i-th element of the list in log time.