Java Arraylist堆实现

Question

I am trying to implement an insert() method for an arraylist-based heap implementation, but whenever I test the method inserting more than one integer where at least one of them is greater than ~4 the program takes forever to run. For example, stating heap.insert(1) and heap.insert(8) together in the main function will take forever to run. Here is my heap class:

import java.util.*;


public class Heap<E extends Comparable<E>> implements HeapAPI<E>
{
    /**
     * A complete tree stored in an array list representing this
     * binary heap
     */
    private ArrayList<E> tree;

    /**
     * Constructs an empty heap
     */
    public Heap()
    {
        tree = new ArrayList<E>();
    }

    public boolean isEmpty()
    {
       return tree.isEmpty();
    }

    public void insert(E obj)
    {
        tree.add(tree.size(), obj);
        siftUp(tree.size() - 1);
    }

    /**
     * siftUp from position k. The key or node value at position
     * may be greater that that of its parent at k/2.
     * @param k position in the heap arraylist
     */
    private void siftUp(int k)
    {
        E v = tree.get(k);
        while (v.compareTo(tree.get(k / 2)) == 1)
        {
            tree.add(k, tree.get(k / 2));
            k = k / 2;
        }
        tree.add(k, v);
    }

    public int size()
    {
       return tree.size();
    }
}

This is the pseudocode given to us by my professor that I modeled the insert() and siftUp() methods from:

The program works for small integers, for example, if I type heap.insert(1) a bunch of times in a row, but never finishes running if I change one of the numbers to a larger integer. No errors are thrown, as it doesn't finish executing when I try to run the program. I'm not sure what's going on. Any help is appreciated.

Answer 1

Let's walk through your code here for a moment:

public void insert(E obj)
{
    tree.add(tree.size(), obj);
    siftUp(tree.size() - 1);
}

private void siftUp(int k)
{
    E v = tree.get(k);
    while (v.compareTo(tree.get(k / 2)) == 1)
    {
        tree.add(k, tree.get(k / 2));
        k = k / 2;
    }
    tree.add(k, v);
}

Now, you do the tree.insert(1) , and everything works fine. Next you call tree.insert(0) . Let's see what happens.

• tree.add(tree.size(), obj) adds 0 as the next item in the array list. So you now have a list that contains [1, 0]
• the code calls siftUp(1)
• in siftUp , the first line gets element 1 from the array list. At this point, v = 1
• at the while statement, the code compares the value, 1, to the value at tree[k/2] . Since k == 1 , you're checking against the root node (ie tree[0] ), which we know is equal to 1.
• Because the new value is smaller than its parent, you copy the parent value to the new item's position. At this point, your array list contains [1, 1] .
• You divide your index, k , by 2, giving 0.
• Now you're back to the while condition. k is equal to 0 and the value at tree[k] is 1, meaning that it's still greater than the value of v.
• You're now in an infinite loop, continually comparing the value at index 0 with your temporary value.

The problem is that your loop has no definite ending condition.

One way to solve the problem is to end the loop if k == 0 .

private void siftUp(int k)
{
    E v = tree.get(k);
    while (k > 0 && v.compareTo(tree.get(k / 2)) == 1)
    {
        tree.add(k, tree.get(k / 2));
        k = k / 2;
    }
    tree.add(k, v);
}

The next trouble you're going to have is that your calculation of the parent node is incorrect. If the root node of the heap is at index 0, then for a node at index ix , its left child is at (ix*2)+1 , and the right child is at (ix*2)+2 . The node's parent is at (ix-1)/2 .

Consider the heap with three nodes:

        0
      1   2

Here the node numbers are the same as the array indexes.

If the root is at index 0, then the left node is at (2 * 0) + 1. The right node is at (2 * 0) + 2. The parent of 1 is (1-1)/2. And the parent of 2 is (2-1)/2. In your code, you parent calculation is k/2 , which would give an incorrect value of 1 for the node labeled 2 in the heap above.