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++.

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());