how to easily sort the same TreeSet few times?

[Victor Fernandez]

Hi!

I am typing my own example of TreeSet.

Basically I have Objects of type Item which have a name of type String and weight of type int. :

    
public class Item implements Comparable {
    private String name;
    private int weight;

    public Item(String name, int weight) {
        this.name = name;
        this.weight = weight;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getWeight() {
        return weight;
    }

    public void setWeight(int weight) {
        this.weight = weight;
    }

    public Item(String name){
        this.name = name;
        this.weight = 0;
    }

    public boolean equals(Object obj) {
        boolean equal = false;
        if((obj instanceof Item) &&
                ((Item)obj).getName().equals(this.name) &&
                ((Item)obj).getWeight()==this.weight){
            equal = true;
        }
        return equal;
    }
    public int hashCode() {
        return (this.name.length() + weight);
    }

    public String toString() {
        return ("\nObject name : " + name + " weight:" + weight);
    }

    public int compareTo(Object obj) {
        int returnValue = -1;
        if(obj instanceof Item){
            returnValue = this.name.compareTo(((Item)obj).getName());
        }
        return returnValue;
    }

}

Well, if I create a TreeSet and I start to add objects they are sorted by name as I specified in the compareTo method.

  
public static void main(String[] args) {
        Item[] items = new Item[10];
        items[0] = new Item("table" , 10 );
        items[1] = new Item("sofa" , 15 );
        items[2] = new Item("tv" , 5 );
        items[3] = new Item("bed" , 20 );
        items[4] = new Item("wardrobe" , 15 );
        items[5] = new Item("shelves" , 2 );
        items[6] = new Item("desk" , 8 );
        items[7] = new Item("chair" , 1 );
        items[8] = new Item("mirror" , 4 );
        items[9] = new Item("mirror" , 4 );

        Set<Item> treeSet = new TreeSet<Item>();
        
        for(Item i : items){
            treeSet.add(i);
        }
        System.out.println(treeSet.toString());
    }
  

At this point my question is: How can I sort the treeSet using for example the weight?
My approach is as follows:
- I have created a comparator and I am trying to sort the TreeSet with it.

   
public class ItemSortByWeight implements Comparator {

    public int compare(Object o1, Object o2) {
        int returnValue = -1;
        if ((o1 instanceof Item) && (o2 instanceof Item)) {
            int o1Weight = ((Item) o1).getWeight();
            int o2Weight = ((Item) o2).getWeight();
            if (o1Weight == o2Weight) {
                returnValue = 0;
            } else if (o1Weight > o2Weight) {
                returnValue = 1;
            }
        }
        return returnValue;


    }

But I don't know how to sort the TreeSet without creating a new one.... I got the items sorted by weight using an array:

  
ItemSortByWeight itemSortByWeight = new ItemSortByWeight();
        Item[] arrayToSort = new Item[treeSet.size()];
        treeSet.toArray(arrayToSort);
        Arrays.sort(arrayToSort, itemSortByWeight);
        
        for (Item i : arrayToSort){
            System.out.println(i.toString());
        } 
 

How can I sort the treeSet without having to create a new one?

Thanks

[Victor Fernandez]

I have read in the API that TreeSet is a Red-black implementation of the NavigableSet interface. And the NavigableSet is a SortedSet extended with navigation methods.

Following with SortedSet the API says that is a Set whose elements are automatically sorted when the SortedSet is created.

Probably the answer to my previous answer could be that : Does not make sense to sort several times a TreeSet! , it is sorted using the sorting we specified in the moment of its creation. And when we add new elements it continues sorted.

Am I right?

Cheers.

[Ken Truitt]

You have your TreeSet<Item> ts and it's sorted based on your compareTo() method; you
now want to sort it some other way? I believe you are right, you can't do it without creating
a new data structure.

[Bob Wheeler]

What about the static Collections method

public static <T> void sort(List<T> list, Comparator<? super T> c)

? Shouldn't that work? Ok, in the end it also uses an array

This implementation dumps the specified list into an array, sorts the array, and iterates over the list resetting each element from the corresponding position in the array. (API)

cheers
Bob

[Victor Fernandez]

What about the static Collections method

public static <T> void sort(List<T> list, Comparator<? super T> c)


? Shouldn't that work? Ok, in the end it also uses an array

This implementation dumps the specified list into an array, sorts the array, and iterates over the list resetting each element from the corresponding position in the array. (API)



cheers

We can sort the treeSet using a List instead of an array. The code could be:

    
List<Item> l =  new ArrayList<Item>(treeSet);
Collections.sort(l, itemSortByWeight);

I am creating an ArrayList from a Collection, in this case a TreeSet.

Cheers