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Asked: May 11, 20262026-05-11T09:56:29+00:00 2026-05-11T09:56:29+00:00

How do I get a PriorityQueue to sort on what I want it to

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How do I get a PriorityQueue to sort on what I want it to sort on?

Also, is there a difference between the offer and add methods?

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  1. 2026-05-11T09:56:30+00:00Added an answer on May 11, 2026 at 9:56 am

    Use the constructor overload which takes a Comparator<? super E> comparator and pass in a comparator which compares in the appropriate way for your sort order. If you give an example of how you want to sort, we can provide some sample code to implement the comparator if you’re not sure. (It’s pretty straightforward though.)

    As has been said elsewhere: offer and add are just different interface method implementations. In the JDK source I’ve got, add calls offer. Although add and offer have potentially different behaviour in general due to the ability for offer to indicate that the value can’t be added due to size limitations, this difference is irrelevant in PriorityQueue which is unbounded.

    Here’s an example of a priority queue sorting by string length:

    // Test.java import java.util.Comparator; import java.util.PriorityQueue;  public class Test {     public static void main(String[] args) {         Comparator<String> comparator = new StringLengthComparator();         PriorityQueue<String> queue = new PriorityQueue<String>(10, comparator);         queue.add("short");         queue.add("very long indeed");         queue.add("medium");         while (queue.size() != 0) {             System.out.println(queue.remove());         }     } }  // StringLengthComparator.java import java.util.Comparator;  public class StringLengthComparator implements Comparator<String> {     @Override     public int compare(String x, String y) {         // Assume neither string is null. Real code should         // probably be more robust         // You could also just return x.length() - y.length(),         // which would be more efficient.         if (x.length() < y.length()) {             return -1;         }         if (x.length() > y.length()) {             return 1;         }         return 0;     } } 

    Here is the output:

    short

    medium

    very long indeed

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