ryBear - 1 month ago 5x
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
{
printHeap();
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
else
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;
heapify(min);

// call heapify
}// end if
printHeap();
}// 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;
}
``````

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)
{
heapify(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

Source (Stackoverflow)