Efficient way to add object to set and order by object property.

Question

I'm trying to build a select menu object. I'm using hibernate criteria to build my object list. I'm trying to take my joined table object and add it to a set to eliminate duplicates. I then want to order the set by a property within the object, once ordered I need to add it to an arraylist and finally add one additional entry to the end of the list.

Here's what I have thus far.

public class Computer {

    private String name;

    //other properties getter/setter
}

Select menu query

public List<Computer> getComputerList() {
    //Hibernate critera query
    List<ComputerConfiguration> results = session.criteraQuery(ComputerConfiguration.class).add(Restrictions.eq("processor", "amd")).list();

    Set<Computer> computerSet = HashSet();

    //Get joined object and add to set to remove duplicates
    for(ComputerConfiguration computerConfiguration : results) {
        computerSet.add(computerConfiguration.getComputer());
    }

    List<Computer> computers = new ArrayList<>(computerSet);
    //Need to somehow order by computer.getName();

    //Lastly add the following
    computers.add("Other");

    //Return computers to select model
    return computers;
}

What is the most efficient way to remove duplicate objects and then order them by the property name? I tried just initially ordering the critera query, however the set didn't appear to hold its position.

Thanks.

Answer 1

Using a HashSet to make sure that the computers are unique (just like you're currently doing it) is fine - assuming that you have implemented hashCode and equals appropriately.

In order to sort the List of computers by name, you can just write

Collections.sort(computers, new Comparator<Computer>()
{
    @Override
    public int compare(Computer c0, Computer c1)
    {
        return c0.getName().compareTo(c1.getName());
    }
});

While it might be tempting to solve this with a TreeSet (because it can sort and remove duplicates in one run), you should carefully think about whether your comparison criterion is consistent with equals as described in the JavaDoc of TreeSet and Comparator:

The ordering imposed by a comparator c on a set of elements S is said to be consistent with equals if and only if c.compare(e1, e2)==0 has the same boolean value as e1.equals(e2) for every e1 and e2 in S.

( http://docs.oracle.com/javase/7/docs/api/java/util/Comparator.html )

The crucial question here is: Can there be two computers that have the same name (as returned by getName()) but should still not be considered as "equal"?

However, if the sorting criterion is consistent with equals, then using TreeSet with the appropriate comparator is feasible.

Answer 2

Use a TreeSet which will sort the elements inside it. You can provide a custom Comparator when creating the set:

Set<Computer> computerSet = new TreeSet<Computer>(new Comparator<Computer>() {
    @Override
    public int compare(Computer computer1, Computer computer2) {
        //write the logic to define which computer is greater than the other...
    }
});

Note that your Computer class won't need to implement equals nor hashCode to be used inside a TreeSet.

Answer 3

You can try to use a SortedSet to actually sort the computer while populating the set : SortedSet

Also, I don't quite understand your line

computers.add("Other")

computers is a Computer object ArrayList you it may not work. So here's my complete solution :

First, make the Computer Object comparable :

public class Computer implements Comparable<Computer> {

    private String name;

    public Computer (String name) { this.name = name; }

    public int compareTo(Computer c) {          
       return name.compareTo(c.name) ;          
   }
}

You can now use a SortedSet like this :

public List<Computer> getComputerList() {
    List<ComputerConfiguration> results = session.criteraQuery(ComputerConfiguration.class).add(Restrictions.eq("processor", "amd")).list();

    SortedSet<Computer> computerSet = new TreeSet<>();
    for(ComputerConfiguration computerConfiguration : results) {
        computerSet.add(computerConfiguration.getComputer());
    }
    computerSet.add(new Computer("Other");

    List<Computer> computers = new ArrayList<>(computerSet);
    return computers;
}