What is the most efficient way (speed-wise) of maintaining a sorted ArrayList in Java?

Question

Which of the following two methods would be the faster one of keeping an ArrayList sorted ?

Inserting the element in its place
Calling push, then call Collections.sort(myList)

I want to have a list of Users(which is a separate class) whose ids must be in ascending order.Here I used Integer as an example in order not to complicate the question — rares985
– rares985, Commented Oct 19, 2015 at 15:33
I think the answer from smonff is the one that makes the most sense: do not build your interfaces around fixed data structures. Instead, select that data structure that gives you what you need; without building additional layers around it. — GhostCat
– GhostCat, Commented Oct 19, 2015 at 15:36
I am kind of new to Java and it was the best approach I could think of.Thank you for your answer — rares985
– rares985, Commented Oct 19, 2015 at 15:39

smonff · Accepted Answer · 2015-10-19 15:31:31Z

3

There is different kinds of collections. You could choose one that keeps your data sorted.

answered Oct 19, 2015 at 15:31

smonff

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1 Comment

assylias Over a year ago

If the op needs random access, I don't think the JDK provides a sorted alternative...

assylias · Accepted Answer · 2015-10-19 15:41:55Z

2

It would probably be more efficient to insert in the right position, for example by using Collections.binarySearch(list, elementToInsert):

int index = Collections.binarySearch(list, element);
if (index < 0) index = - (index + 1);
list.add(index, element);

If you don't need random access and your entries are unique (which seems to be the case based on your comment), you could also use a TreeSet:

NaivigableSet<User> users = new TreeSet<> (Comparator.comparing(User::getId));

The set will always be sorted based on the user ids which must be unique.

edited Oct 19, 2015 at 15:41

answered Oct 19, 2015 at 15:27

assylias

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2 Comments

rares985 Over a year ago

If I will be using your method, my User class will have to implement Comparable<User> and Comparator<User>, right?

assylias Over a year ago

@gusteru In the first example yes, although you can provide your own comparator if required: Collections.binarySearch(list, element, Comparator.comparing(User::getId)). In my second example I have already provided a comparator based on ids so no need for User to be Comparable in that case.

Community · Accepted Answer · 2017-05-23 12:15:34Z

0

If your requirement asks for a continuous sorting after every insert, i think you should try alternate approach than the current one as this in itself is not time efficient. For a elaborate explanation for possible solution you must look at Sorted array list in Java

edited May 23, 2017 at 12:15

CommunityBot

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answered Oct 19, 2015 at 15:34

pradex

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