How can I prove that one algorithm is faster than another in Java

"Ticks"? No. I'd recommend that you run them several times each and compare the average results:

public class AlgorithmDriver {
    public static void main(String [] args) {
        int numTries = 1000000;
        long begTime = System.currentTimeMillis();
        for (int i = 0; i < numTries; ++i) {
            Algorithm.someMethodCall();
        }
        long endTime = System.currentTimeMillis();
        System.out.printf("Total time for %10d tries: %d ms\n", numTries, (endTime-begTime));
    }
}

You probably are asking two different questions:

1. How can you measure the run time of a java implementation (Benchmarking)
2. How can you prove the asymptotic run time of an algorithm

For the first of these I wouldn't use the solutions posted here. They are mostly not quite right. Forst, its probably better to use System.nanoTime than System.currentTimeMillis. Second, you need to use a try catch block. Third, take statistic of running times of your code running many times outside of the metric, so that you can have a more complete picture. Run code that looks vaguely like this many times:

long totalTime = 0;
long startTime = System.nanoTime();
try{
   //method to test
} finally {
   totalTime = System.nanoTime() - startTime;
}

Getting benchmarking correct is hard. For example, you must let your code "warmup"" for a few minutes before testing it. Benchmark early and often, but dont over believe your benchmarks. Particularly small micro benchmarks almost always lie in one way or another.

The second way to interpret your question is about asymptotic run times. The truth is this has almost nothing to do with Java, it is general computer science. Here the question we want to ask is: what curves describe the behavior of the run time of our algorithm in terms of the input size.

The first thing is to understand Big-Oh notation. I'll do my best, but SO doesn't support math notation. O(f(n)) denotes a set of algorithms such that in the limit as n goes to infinity f(n) is within a constant factor of an upper bound on the algorithm run time. Formally, T(n) is in O(f(n)) iff there exists some constant n0 and some constant c such that for all n > n0 c*f(n) >= n. Big Omega is the same thing, except for upper bounds, and big Theta f(n) just means its both big Oh f(n) and big Omega f(n). This is not two hard.

Well, it gets a little more complicated because we can talk about different kinds of run time, ie "average case", best case, and worst case. For example, normall quicksort is O(n^2) in the worst case, but O(n log n) for random lists.

So I skipped over what T(n) means. Basically it is the number of "ticks." Some machine instructions (like reading from memory) take much longer than others (like adding). But, so long as they are only a constant factor apart from each other, we can treat them all as the same for the purposes of big Oh, since it will just change the value of c.

Proving asymptotic bounds isn't that hard. For simple structured programming problems you just count

public int square(int n){
   int sum = 0
   for(int i = 0, i < n, i++){
     sum += n
   }
   return sum
}

In this example we have one instruction each for: initializing sum, initializing i, and returning the value. The loop happens n times and on each time we do a comparison, and addition, and an increment. So we have O(square(n)) = O(3 + 3n) using n0 of 2 and c of 4 we can easily prove this is in O(n). You can always safely simplify big Oh expressions by removing excess constant terms, and by dividing by constant multiples.

When you are faced with a recursive function you have to solve a recurrence relation. If you have a function like T(n) = 2*T(n/2) + O(1) you want to find a closed form solution. You sometimes have to do this by hand, or with a computer algebra system. For this example, using forward substitution, we can see the pattern (in an abuse of notation) T(1) = O(1), T(2) = O(3), T(4) = O(7), T(8) = (15) this looks alot like O(2n - 1), to prove this is the right value:

 T(n) = 2*T(n/2) + 1
 T(n) = 2*(2(n/2) - 1) + 1
 T(n) = 2*(n-1) + 1
 T(n) = 2n - 2 + 1
 T(n) = 2n - 1

As we saw earlier you can simplify O(2n -1) to O(n)

More often though you can use the master theorem which is a mathematical tool for saving you time on this kind of problem. If you check wikipedia you can find the master theorem, which if you plug and play the example above you get the same answer.

For more, check out an algorithms text book like Levitin's "The Design & Analysis of Algorithms"

You could use System.currentTimeMillis() to get start and end times.

long start = System.currentTimeMillis();

// your code

long end = System.currentTimeMillis();
System.out.println( "time: " + (end - start) );