如何在不使用Collections.sort（）的情况下获得Java中的排序列表行为？

Question

I understand that Java does not possess a Sorted List for various conceptual reasons, but consider the case I need to have a collection which is kind of like a Priority Queue but also allows me random access (indexable), in other words, I need a List that follows a particular ordering. I would prefer not to use Collections.sort()

Preferable operation constraints:

retrieve - O(1) (index-based random access)
search - O(log n)
insert - O(log n)
delete - O(log n)

An iterator over the collection should give me all elements in Sorted Order (based on predefined Comparator supplied during instantiation of the data-structure)

I would prefer to use Java's inbuilt library to accomplish this, but feel free to suggest external libraries as well.

EDIT: TreeSet won't do as index based access is difficult, using wrapper collections is also not my best choice as removal would imply I need to remove from both collections.

EDIT2: I was unable to find an implementation and/or documentation for an indexable skip list this seems a bit relevant, can anyone help me find it ? Any comments for or against the data structure proposed is also welcome.

EDIT3: Though this may not be the most perfect answer, I want to add this piece of code that I wrote so that anyone who has similar problems for the need of a sorted list can use this if they find it useful.

Do check for errors (if any), and suggest improvements (especially to the sortedSubList method)

import java.util.ArrayList;
import java.util.Collection;
import java.util.Comparator;

public class SortedList<E> extends ArrayList<E> {
    private final Comparator<? super E> comparator;

    public SortedList(Comparator<? super E> comparator) {
        this.comparator = comparator;
    }

    public SortedList(int initialCapacity, Comparator<? super E> comparator) {
        super(initialCapacity);
        this.comparator = comparator;
    }

    @Override
    public boolean add(E e) {
        if (comparator == null)
            return super.add(e);
        if (e == null)
            throw new NullPointerException();
        int start = 0;
        int end = size() - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (comparator.compare(get(mid), e) == 0) {
                super.add(mid, e);
                return true;
            }
            if (comparator.compare(get(mid), e) < 0) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        super.add(start, e);
        return true;
    }

    @Override
    public boolean contains(Object o) {
        if (comparator == null)
            return super.contains(o);
        if (o == null)
            return false;
        E other = (E) o;
        int start = 0;
        int end = size() - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (comparator.compare(get(mid), other) == 0) {
                return true;
            }
            if (comparator.compare(get(mid), other) < 0) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        return false;
    }

    @Override
    public int indexOf(Object o) {
        if (comparator == null)
            return super.indexOf(o);
        if (o == null)
            throw new NullPointerException();
        E other = (E) o;
        int start = 0;
        int end = size() - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (comparator.compare(get(mid), other) == 0) {
                return mid;
            }
            if (comparator.compare(get(mid), other) < 0) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        return -(start+1);
    }

    @Override
    public void add(int index, E e) {
        throw new UnsupportedOperationException();
    }

    @Override
    public boolean addAll(int index, Collection<? extends E> c) {
        throw new UnsupportedOperationException();
    }

    @Override
    public E set(int index, E e) {
        throw new UnsupportedOperationException();
    }

    public SortedList<E> sortedSubList(int fromIndex, int toIndex) {
        SortedList<E> sl = new SortedList<>(comparator);
        for (int i = fromIndex; i < toIndex; i++)
            sl.add(get(i));
        return sl;
    }
}

Answer 1

It's hard to get O(1) indexing and O(log n) insertion/deletion in the same data structure. O(1) indexing means we can't afford the link-following involved in indexing a tree, list, skip list, or other link-based data structure, while O(log n) modification means we can't afford to shift half the elements of an array on every insertion. I don't know if it's possible to fulfill these requirements simultaneously.

If we relax one of these requirements, things become much easier. For example, O(log n) for all operations can be achieved by an indexable skip list or a self-balancing BST with nodes that keep track of the size of the subtree rooted at the node. Neither of these can be built on top of the skip list or BST in Java's standard library, though, so you'd probably need to install another library or write your own data structure.

O(1) indexing, O(log n) search, and O(n) insert and delete can be done by keeping a sorted ArrayList and using Collections.binarySearch to search for elements or insert/delete positions. You never need to call Collections.sort , but you still need to call the ArrayList's O(n) insert and delete methods. This is probably the easiest option to build on top of Java's built-in tools. Note that with recent Java versions, Collections.sort is an adaptive mergesort that would take O(n) time to sort an array where only the last element is out of sorted order, so you could probably get away with relying on Collections.sort . However, that's an implementation detail that alternative Java implementations don't have to follow.

Answer 2

If your primary goal is O(1) for indexed lookup ( get() ), then you can implement your own class implementing List , backed by an array, using Arrays.binarySearch() .

retrieve: get(int)         - O(1)     - array index
search:   contains(Object) - O(log n) - binarySearch
          indexOf(Object)  - O(log n) - binarySearch
insert:   add(E)           - O(n)     - binarySearch + array shift
delete:   remove(int)      - O(n)     - array shift
          remove(Object)   - O(n)     - binarySearch + array shift

The add(E) method is violating the List definition (append), but is consistent with the Collection definition.

The following methods should throw UnsupportedOperationException :

add(int index, E element)
addAll(int index, Collection<? extends E> c)
set(int index, E element)

If duplicate values are not allowed, which could be a logical restriction, consider also implementing NavigableSet , which is a SortedSet .

Answer 3

Build a custom collection that is backed by an ArrayList and a TreeSet. Delegate the random access to the ArrayList and the search to the TreeSet. Of course this means that every write operation will be very expensive, as it will have to sort the ArrayList every time. But the reads should be very efficient.