I'm baffled that I can't find a quick answer to this. I'm essentially looking for a datastructure in Java which implements the java.util.List
interface, but which stores its members in a sorted order. I know that you can use a normal ArrayList
and use Collections.sort()
on it, but I have a scenario where I am occasionally adding and often retrieving members from my list and I don't want to have to sort it every time I retrieve a member in case a new one has been added. Can anyone point me towards such a thing which exists in the JDK or even 3rd party libraries?
EDIT: The datastructure will need to preserve duplicates.
ANSWER's SUMMARY: I found all of this very interesting and learned a lot. Aioobe in particular deserves mention for his perseverance in trying to achieve my requirements above (mainly a sorted java.util.List implementation which supports duplicates). I have accepted his answer as the most accurate for what I asked and most thought provoking on the implications of what I was looking for even if what I asked wasn't exactly what I needed.
The problem with what I asked for lies in the List interface itself and the concept of optional methods in an interface. To quote the javadoc:
The user of this interface has precise control over where in the list each element is inserted.
Inserting into a sorted list doesn't have precise control over insertion point. Then, you have to think how you will handle some of the methods. Take add
for example:
public boolean add(Object o)
Appends the specified element to the end of this list (optional operation).
You are now left in the uncomfortable situation of either
1) Breaking the contract and implementing a sorted version of add
2) Letting add
add an element to the end of the list, breaking your sorted order
3) Leaving add
out (as its optional) by throwing an UnsupportedOperationException
and implementing another method which adds items in a sorted order.
Option 3 is probably the best, but I find it unsavory having an add method you can't use and another sortedAdd method which isn't in the interface.
Other related solutions (in no particular order):
add(Object obj)
method and bizarrely has a no effect method for add(int index, Object obj)
. General consensus suggests throw new UnsupportedOperationException()
might be a better choice in this scenario.Warning: This class breaks the contract required by List
Here is a "minimal" solution.
class SortedArrayList<T> extends ArrayList<T> {
@SuppressWarnings("unchecked")
public void insertSorted(T value) {
add(value);
Comparable<T> cmp = (Comparable<T>) value;
for (int i = size()-1; i > 0 && cmp.compareTo(get(i-1)) < 0; i--)
Collections.swap(this, i, i-1);
}
}
The insert runs in linear time, but that would be what you would get using an ArrayList anyway (all elements to the right of the inserted element would have to be shifted one way or another).
Inserting something non-comparable results in a ClassCastException. (This is the approach taken by PriorityQueue
as well: A priority queue relying on natural ordering also does not permit insertion of non-comparable objects (doing so may result in ClassCastException).)
List.add
Note that overriding List.add
(or List.addAll
for that matter) to insert elements in a sorted fashion would be a direct violation of the interface specification. What you could do, is to override this method to throw an UnsupportedOperationException
.
From the docs of List.add
:
boolean add(E e)
Appends the specified element to the end of this list (optional operation).
Same reasoning applies for both versions of add
, both versions of addAll
and set
. (All of which are optional operations according to the list interface.)
SortedArrayList<String> test = new SortedArrayList<String>();
test.insertSorted("ddd"); System.out.println(test);
test.insertSorted("aaa"); System.out.println(test);
test.insertSorted("ccc"); System.out.println(test);
test.insertSorted("bbb"); System.out.println(test);
test.insertSorted("eee"); System.out.println(test);
....prints:
[ddd]
[aaa, ddd]
[aaa, ccc, ddd]
[aaa, bbb, ccc, ddd]
[aaa, bbb, ccc, ddd, eee]