Big O: How to determine runtime for a for loop inc

问题:

I have the following algorithm and the runtime complexity is O(N^2) but I want to have a deeper understanding of it rather than just memorizing common runtimes.

What would be the right approach to break it down and analyze it with i+1 in the inner for loop taken into account?

void printunorderedPairs(int[] array) {
    for(int i=0; i<array.length; i++) {
        for(int j=i+1; j<array.length; j++) {
            System.out.println(array[i] + "," + array[j]);
        }
    }
}

EDIT

Asking for how to analyze a specific question

回答1:

What would be the right approach to break it down and analyze it

Take pencil and paper and put down some loops unwraped:

     i        inner loops per i
-------------------------------
     1               length - 1  
     2               length - 2
    ..                       ..  
     k               length - k 
    ..                       ..
length - 1                    1
length                        0

Now, in order to obtain the total time required, let's sum up the inner loops:

 (length - 1) + (length - 2) + ... + (length - k) ... + 1 + 0

It's an arithmetic progression, and its sum is

 ((length - 1) + 0) / 2 * length == length**2 / 2 - length / 2 = O(length**2)

回答2:

Let T = the number of times the inner loop runs.

About half the time, when i<array.length/2, it runs at least array.length/2 times. So, for about array.length/2 outer iterations, the inner loop runs at least array.length/2 times, therefore:

T >= (N/2)*(N/2)
i.e.,
T >= N²/4

This is in O(N²). Also, though, for all array.length outer iterations, the inner loop runs at most array.length times, so:

T <= N*N, i.e.,
T <= N²

This is also in O(N²). Since we have upper and lower bounds on the time that are both in O(N²), we know that T is in O(N²).

NOTE: Technically we only need to upper bound to show that T is in O(N²), but we're usually looking for bounds as tight as we can get. T is actually in Ө(N²).

NOTE ALSO: there's nothing special about using halves above -- any constant proportion will do. These are the general rules in play:

1. Lower bound: If you do at least Ω(N) work at least Ω(N) times, you are doing Ω(N²) work. Ω(N)*Ω(N) = Ω(N²)

2. Upper bound: If you do at most O(N) work at most O(N) times, you are doing O(N²) work. O(N)*O(N) = O(N²)

3. And since we have both, we can use: Ω(N²) ∩ O(N²) = Ө(N²)