Shellsort Interval Question Java

I need to test the efficiency of shellsort when I am using the standard interval size and also while using a non-standard size. The problem I am encountering is when I try to use my non-standard inte开发者_JAVA百科rval.

This is my Shellsort when h is equal to the standard interval size:

    public void shellSort() 
    {
    int inner, outer;
    int temp;
    int h = 1;

    while (h <= length / 3)
    {
      h = h * 3 + 1; 
    }

    while (h > 0) 
    {

      for (outer = h; outer < length; outer++) 
      {
        temp = data[outer];
        inner = outer;

        while (inner > h - 1 && data[inner - h] >= temp) 
        {
          data[inner] = data[inner - h];
          inner -= h;
        }
        data[inner] = temp;
      }

      h = (h - 1) / 3; 

    }

  }

And here is my attempt at using a prime number interval

      private int[]  primes = {0, 1, 3, 7, 13, 31, 97, 211, 503, 1013, 2503, 5171};
      public void shellSort() 
      {
        int inner, outer;
        int temp;
        int count = this.h.length - 1;
        int h = 1;

        h = primes[primes.length - 1] * 2 > length ? primes[primes.length - 1] : primes[primes.length - 2];

        while (h > 0) 
        {         
          for (outer = h; outer < length; outer++) 
          {
            temp = data[outer];
            inner = outer;

            while (inner > h - 1 && data[inner - h] >= temp) 
            {
              data[inner] = data[inner - h];
              inner -= h;
            }
            data[inner] = temp;
          }

          if(count - 1 > 0)           
              h = primes[count - 1];              

        }

      }

I am trying to compare the two based off of real time efficiency , and I can't figure out how to get this prim interval to work.

I'm trying to test:

• Shellsort performs better than O(N^2) with appropriately chosen interval sizes
• The series of interval sizes chosen is important to achieving better than O(N^2) runtime

Thank you for any help.

---

You probably want to decrement the value of count in each iteration of the outer loop. In your code it is still this.h.length-1, which is 11. Therefore, after each iteration of the outer loop you have the if condition count-1 > 0 satisfied, so you set h = this.h[count-1], which I believe is 2503. So, you reenter the loop.

By the way, calling the list of interval sizes h seriously impedes readability. You should call it at least hs.