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 https://stackoverflow.com/a/39020777/56778