Help with heapsort. Pretty new to java

Question

Hello,
I am following a tutorial on heapsort for my java class. I implemented it but the problem is this. The first element in the array i pass in is never changed. Here is the code. What am i doing wrong from this tutorial. What should i do?
public static <T extends Comparable<T>> void heapSort(T[] array)
        heapit(array.length - 1,array);
  public static<T extends Comparable<T>>void swap(int i, int j,T[] array)
      T temp = array[j];
       array[j] = array;
array[i] = temp;
public static <T extends Comparable<T>>void heapit( int end,T[] array )
for ( int i = end / 2; i >= 1; i-- )
fixheap( i, end, array[i],array );
for ( int i = end; i > 1; i-- )
swap( 1, i,array );
fixheap( 1,i - 1 , array[1],array );
public static <T extends Comparable<T>>void fixheap( int root, int end,
T key,T[] array )
int child = 2 * root;
if ( child < end && array[child].compareTo(array[child + 1]) < 0 )
child++;
if ( child <= end && key.compareTo(array[child])< 0 )
array[root] = array[child];
fixheap( child, end, key,array );
else
array[root] = key;

Answer 1

ejp wrote:
It's usual to use 1-based indexing in heapsort algorithms.
So maybe the problem is that you are expecting element[0] to be sorted and it isn't? which it won't be, because of the 1-basing.{noformat}
There's nothing a little math can't cure: for a 1 based algorithm a node at position i has its children at position 2*i and 2*i+1 so for a 0 based indexin scheme the node is at position j+1 and its children are at position 2*(j+1) and 2*(j+1)+1. Because the nodes are stored one position to the left the children of node j are 2*(j+1)-1 and 2*(j+1)+1-1.{noformat}
kind regards,
Jos