Java : Priority Queue | 易学教程

问题

I have a java program which goes like this

public class PriorityQueueExample {

public static void main(String[] args) {
    PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
    pq.add(10);
    pq.add(1);
    pq.add(9);
    pq.add(2);
    pq.add(8);
    pq.add(3);
    pq.add(7);
    pq.add(4);
    pq.add(6);
    pq.add(5);
System.out.println(pq);

}

}

My Question is why does not the priority queue sort them. As per the java specs it implements comparable and maintains the sorting order(natural sorting)

My output of the program is as follows : [1, 2, 3, 4, 5, 9, 7, 10, 6, 8]

回答1:

It is sorted, but internally the elements are stored in a heap. If you call peek(), poll(), or remove(), you will get the right order (and that's how you access queues).

回答2:

Insertion into a priority queue is not enough to sort a list of elements, since it doesn't store them in sorted order; it stores them in the partially sorted heap order. You have to remove the elements in a loop to sort them:

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

回答3:

poll() and remove() will give sorted order not peek() as per java8.

 PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
    pq.add(10);
    pq.add(1);
    pq.add(9);
    pq.add(2);
    pq.add(8);
    pq.add(3);
    pq.add(7);
    pq.add(4);
    pq.add(6);
    pq.add(5);
        // Remove items from the Priority Queue (DEQUEUE)
        while (!pq.isEmpty()) {
          //  System.out.println(pq.remove());
          System.out.println(pq.poll());
        }
Output for poll() & remove():
1 
2
3 
4 
5 
6 
7 
8 
9 
10
output for peek():
1
1
1
1
1
1
1
1
1
1

来源：https://stackoverflow.com/questions/7927213/java-priority-queue

标签

java

collections

priority-queue