Why is there no SortedList in Java?

Question

In Java there are the SortedSet and SortedMap interfaces. Both belong to the Java Collections framework and provide a sorted way to access the elements.

However, in my understanding there is no SortedList in Java. You can use java.util.Collections.sort() to sort a list.

Any idea why it is designed like that?

Answer 1

SortedList in JavaFX

You can find a sorted List implementation in JavaFX, now implemented in OpenJFX: the SortedList class.

Provides a sorted view on an ObservableList.

You can add the OpenJFX libraries to your project. Download from Gluon, or from Oracle. Or you can use a JDK that comes bundled with the OpenJFX libraries, from vendors such as Azul Systems and Bellsoft.

Answer 2

Set and Map are non-linear data structure. List is linear data structure.

---

The tree data structure SortedSet and SortedMap interfaces implements TreeSet and TreeMap respectively using used Red-Black tree implementation algorithm. So it ensure that there are no duplicated items (or keys in case of Map).

• List already maintains an ordered collection and index-based data structure, trees are no index-based data structures.
• Tree by definition cannot contain duplicates.
• In List we can have duplicates, so there is no TreeList(i.e. no SortedList).
• List maintains elements in insertion order. So if we want to sort the list we have to use java.util.Collections.sort(). It sorts the specified list into ascending order, according to the natural ordering of its elements.
• Array, List are not automatically sorted (can be sorted manually).

Answer 3

As discussed already, SortedList (aka SortedSequence) would break the expectations of users of a List that add(x) adds x to the list end, and add(i,x) adds x at position i, etc.

However these are optional operations of List according to the Java SE8 documentation, so there is no guarantee about them in any case: https://docs.oracle.com/javase/8/docs/api/java/util/List.html.

SortedList would be a useful collection if you wish to have a consistent sorted order imposed on a list throughout program processing, it also will enable faster search operations such as contains, indexOf, lastIndexOf, getCount, etc, although insertion will be slower. It can also be used to support the definition of a SortedOrderedSet (a unique-element sorted sequence, constructed using a HashSet and a SortedSequence).

Implementations of SortedSequence and SortedOrderedSet are provided as part of the AgileUML libraries: https://github.com/eclipse/agileuml/tree/master/libraries

Answer 4

List iterators guarantee first and foremost that you get the list's elements in the internal order of the list (aka. insertion order). More specifically it is in the order you've inserted the elements or on how you've manipulated the list. Sorting can be seen as a manipulation of the data structure, and there are several ways to sort the list.

I'll order the ways in the order of usefulness as I personally see it:

1. Consider using Set or Bag collections instead

NOTE: I put this option at the top because this is what you normally want to do anyway.

A sorted set automatically sorts the collection at insertion, meaning that it does the sorting while you add elements into the collection. It also means you don't need to manually sort it.

Furthermore if you are sure that you don't need to worry about (or have) duplicate elements then you can use the TreeSet<T> instead. It implements SortedSet and NavigableSet interfaces and works as you'd probably expect from a list:

TreeSet<String> set = new TreeSet<String>();
set.add("lol");
set.add("cat");
// automatically sorts natural order when adding

for (String s : set) {
    System.out.println(s);
}
// Prints out "cat" and "lol"

If you don't want the natural ordering you can use the constructor parameter that takes a Comparator<T>.

Alternatively, you can use Multisets (also known as Bags), that is a Set that allows duplicate elements, instead and there are third-party implementations of them. Most notably from the Guava libraries there is a TreeMultiset, that works a lot like the TreeSet.

2. Sort your list with Collections.sort()

As mentioned above, sorting of Lists is a manipulation of the data structure. So for situations where you need "one source of truth" that will be sorted in a variety of ways then sorting it manually is the way to go.

You can sort your list with the java.util.Collections.sort() method. Here is a code sample on how:

List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");

Collections.sort(strings);
for (String s : strings) {
    System.out.println(s);
}
// Prints out "cat" and "lol"

Using comparators

One clear benefit is that you may use Comparator in the sort method. Java also provides some implementations for the Comparator such as the Collator which is useful for locale sensitive sorting strings. Here is one example:

Collator usCollator = Collator.getInstance(Locale.US);
usCollator.setStrength(Collator.PRIMARY); // ignores casing

Collections.sort(strings, usCollator);

In Java 8+, more simply call List#sort.

strings.sort( usCollator ) ;

Sorting in concurrent environments

Do note though that using the sort method is not friendly in concurrent environments, since the collection instance will be manipulated, and you should consider using immutable collections instead. This is something Guava provides in the Ordering class and is a simple one-liner:

List<string> sorted = Ordering.natural().sortedCopy(strings);

3. Wrap your list with java.util.PriorityQueue

Though there is no sorted list in Java there is however a sorted queue which would probably work just as well for you. It is the java.util.PriorityQueue class.

Nico Haase linked in the comments to a related question that also answers this.

In a sorted collection you most likely don't want to manipulate the internal data structure which is why PriorityQueue doesn't implement the List interface (because that would give you direct access to its elements).

Caveat on the PriorityQueue iterator

The PriorityQueue class implements the Iterable<E> and Collection<E> interfaces so it can be iterated as usual. However, the iterator is not guaranteed to return elements in the sorted order. Instead (as Alderath points out in the comments) you need to poll() the queue until empty.

Note that you can convert a list to a priority queue via the constructor that takes any collection:

List<String> strings = new ArrayList<String>()
strings.add("lol");
strings.add("cat");

PriorityQueue<String> sortedStrings = new PriorityQueue(strings);
while(!sortedStrings.isEmpty()) {
    System.out.println(sortedStrings.poll());
}
// Prints out "cat" and "lol"

4. Write your own SortedList class

NOTE: You shouldn't have to do this.

You can write your own List class that sorts each time you add a new element. This can get rather computation heavy depending on your implementation and is pointless, unless you want to do it as an exercise, because of two main reasons:

1. It breaks the contract that List<E> interface has because the add methods should ensure that the element will reside in the index that the user specifies.
2. Why reinvent the wheel? You should be using the TreeSet or Multisets instead as pointed out in the first point above.

However, if you want to do it as an exercise here is a code sample to get you started, it uses the AbstractList abstract class:

public class SortedList<E> extends AbstractList<E> {

    private ArrayList<E> internalList = new ArrayList<E>();

    // Note that add(E e) in AbstractList is calling this one
    @Override 
    public void add(int position, E e) {
        internalList.add(e);
        Collections.sort(internalList, null);
    }

    @Override
    public E get(int i) {
        return internalList.get(i);
    }

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

}

Note that if you haven't overridden the methods you need, then the default implementations from AbstractList will throw UnsupportedOperationExceptions.

Answer 5

Computational speaking, in simple lists, it is more expensive to insert ordered than insert without any order and then order after you are done inserting.

Answer 6

We have Collections.sort(arr) method which can help to sort ArrayList arr. to get sorted in desc manner we can use Collections.sort(arr, Collections.reverseOrder())

Answer 7

In case you are looking for a way to sort elements, but also be able to access them by index in an efficient way, you can do the following:

1. Use a random access list for storage (e.g. ArrayList)
2. Make sure it is always sorted

Then to add or remove an element you can use Collections.binarySearch to get the insertion / removal index. Since your list implements random access, you can efficiently modify the list with the determined index.

Example:

/**
 * @deprecated
 *      Only for demonstration purposes. Implementation is incomplete and does not 
 *      handle invalid arguments.
 */
@Deprecated
public class SortingList<E extends Comparable<E>> {
    private ArrayList<E> delegate;
    
    public SortingList() {
        delegate = new ArrayList<>();
    }
    
    public void add(E e) {
        int insertionIndex = Collections.binarySearch(delegate, e);
        
        // < 0 if element is not in the list, see Collections.binarySearch
        if (insertionIndex < 0) {
            insertionIndex = -(insertionIndex + 1);
        }
        else {
            // Insertion index is index of existing element, to add new element 
            // behind it increase index
            insertionIndex++;
        }
        
        delegate.add(insertionIndex, e);
    }
    
    public void remove(E e) {
        int index = Collections.binarySearch(delegate, e);
        delegate.remove(index);
    }
    
    public E get(int index) {
        return delegate.get(index);
    }
}

(See a more complete implementation in this answer)

Answer 8

Because the concept of a List is incompatible with the concept of an automatically sorted collection. The point of a List is that after calling list.add(7, elem), a call to list.get(7) will return elem. With an auto-sorted list, the element could end up in an arbitrary position.

Answer 9

https://github.com/geniot/indexed-tree-map

Consider using indexed-tree-map . It's an enhanced JDK's TreeSet that provides access to element by index and finding the index of an element without iteration or hidden underlying lists that back up the tree. The algorithm is based on updating weights of changed nodes every time there is a change.

Answer 10

I think all the above do not answer this question due to following reasons,

1. Since same functionality can be achieved by using other collections such as TreeSet, Collections, PriorityQueue..etc (but this is an alternative which will also impose their constraints i.e. Set will remove duplicate elements. Simply saying even if it does not impose any constraint, it does not answer the question why SortedList was not created by java community)
2. Since List elements do not implements compare/equals methods (This holds true for Set & Map also where in general items do not implement Comparable interface but when we need these items to be in sorted order & want to use TreeSet/TreeMap,items should implement Comparable interface)
3. Since List uses indexing & due to sorting it won't work (This can be easily handled introducing intermediate interface/abstract class)

but none has told the exact reason behind it & as I believe these kind of questions can be best answered by java community itself as it will have only one & specific answer but let me try my best to answer this as following,

As we know sorting is an expensive operation and there is a basic difference between List & Set/Map that List can have duplicates but Set/Map can not. This is the core reason why we have got a default implementation for Set/Map in form of TreeSet/TreeMap. Internally this is a Red Black Tree with every operation (insert/delete/search) having the complexity of O(log N) where due to duplicates List could not fit in this data storage structure.

Now the question arises we could also choose a default sorting method for List also like MergeSort which is used by Collections.sort(list) method with the complexity of O(N log N). Community did not do this deliberately since we do have multiple choices for sorting algorithms for non distinct elements like QuickSort, ShellSort, RadixSort...etc. In future there can be more. Also sometimes same sorting algorithm performs differently depending on the data to be sorted. Therefore they wanted to keep this option open and left this on us to choose. This was not the case with Set/Map since O(log N) is the best sorting complexity.

Answer 11

Since all lists are already "sorted" by the order the items were added (FIFO ordering), you can "resort" them with another order, including the natural ordering of elements, using java.util.Collections.sort().

EDIT:

Lists as data structures are based in what is interesting is the ordering in which the items where inserted.

Sets do not have that information.

If you want to order by adding time, use List. If you want to order by other criteria, use SortedSet.

Answer 12

For any newcomers, as of April 2015, Android now has a SortedList class in the support library, designed specifically to work with RecyclerView. Here's the blog post about it.

Answer 13

First line in the List API says it is an ordered collection (also known as a sequence). If you sort the list you can't maintain the order, so there is no TreeList in Java.
As API says Java List got inspired from Sequence and see the sequence properties http://en.wikipedia.org/wiki/Sequence_(mathematics)

It doesn't mean that you can't sort the list, but Java strict to his definition and doesn't provide sorted versions of lists by default.

Answer 14

Another point is the time complexity of insert operations. For a list insert, one expects a complexity of O(1). But this could not be guaranteed with a sorted list.

And the most important point is that lists assume nothing about their elements. For example, you can make lists of things that do not implement equals or compare.

Answer 15

Think of it like this: the List interface has methods like add(int index, E element), set(int index, E element). The contract is that once you added an element at position X you will find it there unless you add or remove elements before it.

If any list implementation would store elements in some order other than based on the index, the above list methods would make no sense.