I have priority queue in Java of Integers:
PriorityQueue<Integer> pq= new PriorityQueue<Integer>();
When I call pq.poll() I get the minimum element.
Question: how to change the code to get the maximum element?
This question is related to
java
collections
priority-queue
PriorityQueue<Integer> pq = new PriorityQueue<Integer> (
new Comparator<Integer> () {
public int compare(Integer a, Integer b) {
return b - a;
}
}
);
The elements of the priority queue are ordered according to their natural ordering, or by a Comparator provided at queue construction time.
The Comparator should override the compare method.
int compare(T o1, T o2)
Default compare method returns a negative integer, zero, or a positive integer as the first argument is less than, equal to, or greater than the second.
The Default PriorityQueue provided by Java is Min-Heap, If you want a max heap following is the code
public class Sample {
public static void main(String[] args) {
PriorityQueue<Integer> q = new PriorityQueue<Integer>(new Comparator<Integer>() {
public int compare(Integer lhs, Integer rhs) {
if(lhs<rhs) return +1;
if(lhs>rhs) return -1;
return 0;
}
});
q.add(13);
q.add(4);q.add(14);q.add(-4);q.add(1);
while (!q.isEmpty()) {
System.out.println(q.poll());
}
}
}
Reference :https://docs.oracle.com/javase/7/docs/api/java/util/PriorityQueue.html#comparator()
This can be achieved by the below code in Java 8 which has introduced a constructor which only takes a comparator.
PriorityQueue<Integer> maxPriorityQ = new PriorityQueue<Integer>(Collections.reverseOrder());
Using lamda, just multiple the result with -1 to get max priority queue.
PriorityQueue<> q = new PriorityQueue<Integer>(
(a,b) -> -1 * Integer.compare(a, b)
);
In Java 8+ you can create a max priority queue via one of these methods:
Method 1:
PriorityQueue<Integer> maxPQ = new PriorityQueue<>(Collections.reverseOrder());
Method 2:
PriorityQueue<Integer> maxPQ = new PriorityQueue<>((a,b) -> b - a);
Method 3:
PriorityQueue<Integer> maxPQ = new PriorityQueue<>((a,b) -> b.compareTo(a));
You can use MinMaxPriorityQueue (it's a part of the Guava library):
here's the documentation. Instead of poll(), you need to call the pollLast() method.
Here is a sample Max-Heap in Java :
PriorityQueue<Integer> pq1= new PriorityQueue<Integer>(10, new Comparator<Integer>() {
public int compare(Integer x, Integer y) {
if (x < y) return 1;
if (x > y) return -1;
return 0;
}
});
pq1.add(5);
pq1.add(10);
pq1.add(-1);
System.out.println("Peek: "+pq1.peek());
The output will be 10
You can use lambda expression since Java 8.
The following code will print 10, the larger.
// There is overflow problem when using simple lambda as comparator, as pointed out by ???? ?????.
// PriorityQueue<Integer> pq = new PriorityQueue<>((x, y) -> y - x);
PriorityQueue<Integer> pq =new PriorityQueue<>((x, y) -> Integer.compare(y, x));
pq.add(10);
pq.add(5);
System.out.println(pq.peek());
The lambda function will take two Integers as input parameters, subtract them from each other, and return the arithmetic result. The lambda function implements the Functional Interface, Comparator<T>. (This is used in place, as opposed to an anonymous class or a discrete implementation.)
This can be achieved by using
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(Collections.reverseOrder());
You can provide a custom Comparator object that ranks elements in the reverse order:
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(defaultSize, new Comparator<Integer>() {
public int compare(Integer lhs, Integer rhs) {
if (lhs < rhs) return +1;
if (lhs.equals(rhs)) return 0;
return -1;
}
});
Now, the priority queue will reverse all its comparisons, so you will get the maximum element rather than the minimum element.
Hope this helps!
Change PriorityQueue to MAX PriorityQueue Method 1 : Queue pq = new PriorityQueue<>(Collections.reverseOrder()); Method 2 : Queue pq1 = new PriorityQueue<>((a, b) -> b - a); Let's look at few Examples:
public class Example1 {
public static void main(String[] args) {
List<Integer> ints = Arrays.asList(222, 555, 666, 333, 111, 888, 777, 444);
Queue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
pq.addAll(ints);
System.out.println("Priority Queue => " + pq);
System.out.println("Max element in the list => " + pq.peek());
System.out.println("......................");
// another way
Queue<Integer> pq1 = new PriorityQueue<>((a, b) -> b - a);
pq1.addAll(ints);
System.out.println("Priority Queue => " + pq1);
System.out.println("Max element in the list => " + pq1.peek());
/* OUTPUT
Priority Queue => [888, 444, 777, 333, 111, 555, 666, 222]
Max element in the list => 888
......................
Priority Queue => [888, 444, 777, 333, 111, 555, 666, 222]
Max element in the list => 888
*/
}
}
Let's take a famous interview Problem : Kth Largest Element in an Array using PriorityQueue
public class KthLargestElement_1{
public static void main(String[] args) {
List<Integer> ints = Arrays.asList(222, 555, 666, 333, 111, 888, 777, 444);
int k = 3;
Queue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
pq.addAll(ints);
System.out.println("Priority Queue => " + pq);
System.out.println("Max element in the list => " + pq.peek());
while (--k > 0) {
pq.poll();
} // while
System.out.println("Third largest => " + pq.peek());
/*
Priority Queue => [888, 444, 777, 333, 111, 555, 666, 222]
Max element in the list => 888
Third largest => 666
*/
}
}
Another way :
public class KthLargestElement_2 {
public static void main(String[] args) {
List<Integer> ints = Arrays.asList(222, 555, 666, 333, 111, 888, 777, 444);
int k = 3;
Queue<Integer> pq1 = new PriorityQueue<>((a, b) -> b - a);
pq1.addAll(ints);
System.out.println("Priority Queue => " + pq1);
System.out.println("Max element in the list => " + pq1.peek());
while (--k > 0) {
pq1.poll();
} // while
System.out.println("Third largest => " + pq1.peek());
/*
Priority Queue => [888, 444, 777, 333, 111, 555, 666, 222]
Max element in the list => 888
Third largest => 666
*/
}
}
As we can see, both are giving the same result.
You can try pushing elements with reverse sign. Eg: To add a=2 & b=5 and then poll b=5.
PriorityQueue<Integer> pq = new PriorityQueue<>();
pq.add(-a);
pq.add(-b);
System.out.print(-pq.poll());
Once you poll the head of the queue, reverse the sign for your usage. This will print 5 (larger element). Can be used in naive implementations. Definitely not a reliable fix. I don't recommend it.
PriorityQueue<Integer> lowers = new PriorityQueue<>((o1, o2) -> -1 * o1.compareTo(o2));
We can do this by creating our CustomComparator class that implements Comparator interface and overriding its compare method. Below is the code for the same :
import java.util.PriorityQueue;
import java.util.Comparator;
public class Main
{
public static void main(String[] args) {
PriorityQueue<Integer> nums = new PriorityQueue<>(new CustomComparator());
nums.offer(21);
nums.offer(1);
nums.offer(8);
nums.offer(2);
nums.offer(-4);
System.out.println(nums.peek());
}
}
class CustomComparator implements Comparator<Integer>{
@Override
public int compare(Integer n1, Integer n2){
int val = n1.compareTo(n2);
if(val > 0)
return -1;
else if(val < 0)
return 1;
else
return 0;
}
}
You can try something like:
PriorityQueue<Integer> pq = new PriorityQueue<>((x, y) -> -1 * Integer.compare(x, y));
Which works for any other base comparison function you might have.
I just ran a Monte-Carlo simulation on both comparators on double heap sort min max and they both came to the same result:
These are the max comparators I have used:
(A) Collections built-in comparator
PriorityQueue<Integer> heapLow = new PriorityQueue<Integer>(Collections.reverseOrder());
(B) Custom comparator
PriorityQueue<Integer> heapLow = new PriorityQueue<Integer>(new Comparator<Integer>() {
int compare(Integer lhs, Integer rhs) {
if (rhs > lhs) return +1;
if (rhs < lhs) return -1;
return 0;
}
});
Source: Stackoverflow.com