How to use pair in a priority queue and then return the value using the key as the priority

Question

So I want to use the smallest key as the priority and then return the VALUE of that corresponding key:

import javafx.util.Pair;
import java.util.PriorityQueue;

public class Test
{
    public static void main (String[] args)
    {
        int n = 5;

        PriorityQueue <Pair <Integer,Integer> > l = new PriorityQueue <Pair <Integer,Integer> > (n);

        l.add(new Pair <> (1, 90));
        l.add(new Pair <> (7, 54));
        l.add(new Pair <> (2, 99));
        l.add(new Pair <> (4, 88));
        l.add(new Pair <> (9, 89));

        System.out.println(l.poll().getValue()); 
    }
}

The output Im looking for is 90 because 1 is the smallest key. Its fine even if the value is used as the priority and the key is returned because I can just swap the data if neccessary. I want to display key/value using value/key as a priority (minimum value in this case). I do not know how this can be done in this scenario. This works fine in C++.

Cristalcristate

Try using Comparator.comparing(Pair::getValue). – Ebert

can you please elaborate a little more. Or just post an modification. It'd help because I'm learning and uncomfortable with collections. Thanks – Cristalcristate

There is a constructor public PriorityQueue(int initialCapacity, Comparator<? super E> comparator) which accepts a comparator as second argument. The comparator describes how to sort. – Android

Answer 1

You need to use Comparator which will be used to order this priority queue.

Use Comparator.comparing() and pass Method reference of comparing parameter when create the PriorityQueue

PriorityQueue<Pair<Integer,Integer> > pq=
                new PriorityQueue<Pair<Integer,Integer>>(n, Comparator.comparing(Pair::getKey));

Or

You can use lambda expression

PriorityQueue<Pair<Integer,Integer> > pq=
                    new PriorityQueue<Pair<Integer,Integer>>(n,(a,b) -> a.getKey() - b.getKey());

Takashi

Alright I was able to figure it out by using the above constructor and then using poll().getValue() to return the value – Cristalcristate

Answer 2

Alternative, use custom Pair class that implements Comparable.

class Pair implements Comparable<Pair> {
    Integer value;
    Integer index;

    public Pair(Integer value, Integer index) {
        this.value = value;
        this.index = index;
    }

    @Override
    public int compareTo(Pair o) {
        return value - o.value;
    }
}

and then usage

// Add elements
queue.add(new Pair(valueKey, index));
// do it for all your elements

final Pair min = queue.poll();
Integer index = min.index;
Integer value = min.value;

// Do with them what you want.

I used PriorityQueue in leetcode challenge. https://leetcode.com/problems/merge-k-sorted-lists/discuss/630580/Using-PriorityQueue-in-Java

And this is full example https://github.com/yan-khonski-it/leetcode/blob/master/src/main/java/com/yk/training/leetcode/merge_sorted_lists/PriorityQueueSolution.java

Lyte

More traditional approach. I like it! – Cassandracassandre

Answer 3

I prefer to use Map.Entry in most cases for this purpose. Other time a custom Pair class with Comparator in PriorityQueue constructor.

import java.util.PriorityQueue;
import java.util.Map;
import java.util.HashMap;

public class Test
{
    public static void main (String[] args)
    {
        int n = 5;

        PriorityQueue <Map.Entry<Integer,Integer>> l = new PriorityQueue <> (n,(a,b)->Integer.compare(a.getKey(),b.getKey()));
        Map<Integer,Integer> map = new HashMap<>();
        map.put(1,90);
        map.put(7,54);
        map.put(2,99);
        map.put(4,88);
        map.put(9,89);
        l.addAll(map.entrySet());
        System.out.println(l.poll().getValue()); 
    }
}

Buckboard

Answer 4

This worked for me

val pq = PriorityQueue<Pair<Int, Int>>(compareBy { it.first })

Polyneuritis