Java Max heap __ algorithm

public class Heapbig {public static void main (string[] args) {//TODO automatically generated method stubs int[] A = {10, 8, 11, 8, 14, 9, 4,
		1, 17};
		Heapbig HB = new Heapbig ();
		Hb.heapsort (a);
	for (int x:a) System.out.print (x + ""); For subscript I, its parent node is labeled (i + 1)/2-1, the left child node is labeled I * 2 + 1, the right child node is labeled I * 2 + 2//maintenance maximum heap public void maxheapify (int[] A, int i
		, int length) {int L = left (i);
		int r = Right (i);
		int Max;
		if (l < length && a[l] > A[i]) {max = L;
		else {max = i;
		} if (r < length && A[r] > A[max]) {max = R;
			} if (max!= i) {swap (A, I, max); The value of node Max becomes the value of the smaller original node I, so the subtree with node Max as the root node may violate the nature of the maximum heap//node I becomes node Max, so recursively calls Maxheapify (A, max, length) maxheapify
		, length);
	}///Find the Child left node ordinal of node I public int left (int i) {return I * 2 + 1;
	//Find node I's Sub right node ordinal public int right (int i) {return I * 2 + 2;
		public void swap (int a[], int i, int j) {int temp = A[i];
		A[i] = A[j];
	A[J] = temp; //Establish maximum heap public void Buildmaxheap (int[] a) {for (int i = A.LENGTH/2 + 1; I >= 0; i) {maxheapify (A, I, a.length); The maximum element of the heap sort///whole array is always the first in the largest heap, that is, a[0]. At this point, if we take the first element, put it in the last position in the array, and then maintain the maximum heap maxheapify for the first element, and so on, we can complete the order of the entire array public void Heapsort (int[] a) {//
		First, the maximum heap is established, and the first element is the maximum value buildmaxheap (a);
		int length = A.length; 
			for (int i = a.length-1 i >= 1;-i) {swap (A, I, 0);//move the first largest element to the back, and ignore it in the process of maxheapify by reducing length--length;
		Maxheapify (A, 0, length);
 }
	}
}