max heap python

1import java.util.PriorityQueue;
2
3public class MaxHeapWithPriorityQueue {
4
5    public static void main(String args[]) {
6        // create priority queue
7        PriorityQueue<Integer> prq = new PriorityQueue<>(Comparator.reverseOrder());
8
9        // insert values in the queue
10        prq.add(6);
11        prq.add(9);
12        prq.add(5);
13        prq.add(64);
14        prq.add(6);
15
16        //print values
17        while (!prq.isEmpty()) {
18            System.out.print(prq.poll()+" ");
19        }
20    }
21
22}

1import heapq
2
3Since the built in heapq library is a minheap, multiply your values by -1
4and it will function as a max heap. Just remeber that all your numbers 
5have been inverted.

1def min_heapify(A,k):
2    l = left(k)
3    r = right(k)
4    if l < len(A) and A[l] < A[k]:
5        smallest = l
6    else:
7        smallest = k
8    if r < len(A) and A[r] < A[smallest]:
9        smallest = r
10    if smallest != k:
11        A[k], A[smallest] = A[smallest], A[k]
12        min_heapify(A, smallest)
13
14def left(k):
15    return 2 * k + 1
16
17def right(k):
18    return 2 * k + 2
19
20def build_min_heap(A):
21    n = int((len(A)//2)-1)
22    for k in range(n, -1, -1):
23        min_heapify(A,k)
24
25A = [3,9,2,1,4,5]
26build_min_heap(A)
27print(A)
28

1def max_heapify(A,k):
2    l = left(k)
3    r = right(k)
4    if l < len(A) and A[l] > A[k]:
5        largest = l
6    else:
7        largest = k
8    if r < len(A) and A[r] > A[largest]:
9        largest = r
10    if largest != k:
11        A[k], A[largest] = A[largest], A[k]
12        max_heapify(A, largest)
13
14def left(k):
15    return 2 * k + 1
16
17def right(i):
18    return 2 * k + 2
19
20def build_max_heap(A):
21    n = int((len(A)//2)-1)
22    for k in range(n, -1, -1):
23        max_heapify(A,k)
24
25A = [3,9,2,1,4,5]
26build_max_heap(A)
27print(A)