Complexity of bubble sort
Strictly answering your question about what the \$Big-O\$ run-time complexity,

public class TempBubbleSort 
{

static Integer[] myArray = {10,9,8,7,6,5,4,3,2,1};

static int counter = 0;

public static void main(String[] args)
{
    // **This is O(n), as we're touching every point in your array.
    for(int anchor=0; anchor&lt;myArray.length; anchor++)
    {
        // **This is O(n-1), but since we don't care about constants as n grows, we consider this as O(n)
        for(int compare=anchor+1; compare&lt;myArray.length; compare++)
        {
            // **Everything in here then is a bunch of constant operations so we'll just ignore them.
            counter++;
            // sort ascending
            if(myArray[anchor] &gt; myArray[compare])
            {
                int tmp = myArray[compare];
                myArray[compare] = myArray[anchor];
                myArray[anchor] = tmp;
            }
        }
    }
    // **So the total of the brunt of your work would be O(n)*O(n) or O(n^2), because for every element in your array n, you touch every other element in your array, so you can look at it as touching n things in your array, n times.

    System.out.println("Comparision Count : "+counter);

    // **In case you're curious, this is also O(n)
    for(int i : myArray)
        System.out.println(i);

    // **Which would bring the grand total to O(n^2+n), again with Big O notation, we only care about what happens as n continues to grow, and since n^2 grows faster than n, as our n gets super big, the second term n becomes insignificant, so we look at this overall total as O(n^2). In fact, for any polynomial time complexity, you can safely drop all terms that are not your largest degree.
}


}



In closing, its usually easy to tell the time-complexity of simple iterative functions like this by the number of nested for loops you have, which can be a good rule of thumb for a nice approximate guess for time complexity in a pinch.