Shell Sort - Printing Each step in the algorithm

Question

i have this shell sort algorithm, and i want to print out each step it takes to sort the array of numbers... i just am having trouble finding the proper spot to insert the print(a); statement
public class ShellSort
{ public static void main(String[] args)
{ int[] a = { 59,20,17,13,28,14,23,83,36,98,11,70,65,41,42,15 };
sort(a);
print(a);
public static void print(int[] a)
{ for (int i=0; i
System.out.print(a +" ");
System.out.println();
public static void sort(int[] a)
{ int d=1, j, n=a.length; // step 1
while (9*d
d = 3*d + 1; // step 3
while (d>0) // step 4
{ for (int i=d; i
{ int temp=a ;
j=i;
while (j>=d && a[j-d]>temp)
{ a[j] = a[j-d];
j -= d;
a[j] = temp;
d /= 3; // step 6
}any suggestions or hints???
thanks,
scott.

Answer 1

i'm sorry about the readability...
//  Scott Garshwiler
//  Example 13.2 taken from page 244 of Schaum's Outline
//  Testing the Shell Sort
//  Everything works  cost= theta(n^1.5)
public class ShellSort
{ public static void main(String[] args)
  { int[] a = { 59,20,17,13,28,14,23,83,36,98,11,70,65,41,42,15 };
    sort(a);
    print(a);
  public static void print(int[] a)
  { for (int i=0; i<a.length; i++)
      System.out.print(a[i]+" ");
    System.out.println();
  public static void sort(int[] a)
  { int d=1, j, n=a.length; // step 1
    while (9*d<n) // step 2
      d = 3*d + 1; // step 3
    while (d>0) // step 4
    { for (int i=d; i<n; i++) // step 5
      { int temp=a;
j=i;
while (j>=d && a[j-d]>temp)
{ a[j] = a[j-d];
j -= d;
a[j] = temp;
d /= 3; // step 6