ryBear ryBear - 1 year ago 264
Java Question

Max Heapify issue

I have been trying to debug this code for hours. I don't know why it is not rearranging the terms. I have tried everything I can think of. Can someone help? Thanks.

public void heapify(int i) // utility routine to percolate down from index i
int left, r, min;
Process tmp;

left = lchild(i); // left child
r = rchild(i); // right child

if(left < size() && A[left].compareTo(A[i])<0) // find smallest child
min = left; // save index of smaller child
min = i;

if(r < size() && A[r].compareTo(A[min])<0)
min = r; // save index of smaller child

if(min != i) // swap and percolate, if necessary
tmp = A[i]; // exchange values at two indices
A[i] = A[min];
A[min] = tmp;

// call heapify
}// end if
}// end method heapify

private int lchild(int i) {
return 2 * i + 1;

private int rchild(int i) {
return 2 * i + 2;

Even when I call heapify on every element of the heap it doesn't work :/
Here is the compareTo. It is supposed to arrange max heap using priority first then if there is a tie it goes to a unique time arrived value.

public int compareTo(Process o) {
int val;
if (this.priority > o.getPriority()) {
val = -1;
} else if (this.priority == o.getPriority()) {
if (this.arrivalTime < o.getArrivalTime()) { //Earlier time
val = -1;
} else {
val = 1;
} else {
val = 1;

return val;

Answer Source

The fastest known way to organize an array into a heap is called Floyd's Algorithm. You start at the middle of the array and move towards the root, sifting each item down as required. In your case:

for (int i = size()/2; i >= 0; --i)

You should be able to call the heapify function that you supplied.

To see how this works, take a look at http://stackoverflow.com/a/39020777/56778

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