How is the Java priority Queue supposed to work? [

Question

This question already has answers here:

PriorityQueue.toString wrong element order (4 answers)

Closed 3 years ago.

Short story, I'm implementing a graph and now I'm working on the Kruskal, I need a priority queue. My definition of a priority queue is that the element with the smallest key would come first? Is this wrong? Because when I insert the weighted edges(or numbers) in the queue they don't end up sorted.

PriorityQueue<Integer> tja = new PriorityQueue<Integer>(); 
tja.add(55);
tja.add(99); 
tja.add(1); 
tja.add(102);
tja.add(54);
tja.add(51);
System.out.println(tja);

That would print out this; [1, 54, 51, 102, 99, 55]. This is not sorted like I want them to be! And yes I made a comperator that goes into the priority queue that extracts the number from the edge object and compares based on that int. So this should work, or have I just completely misunderstood the entire concept of how this data structure works?

Answer 1

System.out.println is invoking the toString() method, which is using the iterator, which is not guaranteed to respect the natural ordering. From the docs: "The Iterator provided in method iterator() is not guaranteed to traverse the elements of the priority queue in any particular order."

Answer 2

Are you familiar with functioning of Binary heaps? If not please go through min heap and max heap structures. PriorityQueue is implemented on heaps. PriorityQueue doesn't sort the items in increasing order, but it does the heap sort on it.

Go through the link: Priority Queue

The output you get is correct.

Answer 3

I have no experience with PriorityQueue in Java but it looks like the priority thing is not integrated into iterator() or toString() (which uses iterator()).

If you do:

    while (tja.size()>0)
        System.out.println(tja.remove());

You get the proper results.