如何确定看起来像这样的东西的大O：（x -1）+（x-2）+（x-3）…（x-x）

Question

I'm trying to brush up on my big o calculations. If I have function that shifts all of the items to the right of 'i' 2 spaces I have a formula that looks something like:

(n -1) + (n - 2) + (n - 3) ... (n - n)

Where the first iteration I have to move (x-1) items, the second (x-2) items, and so on... the method:

int[] s = {1,2,3,4, , }

public static char[] moveStringDownTwoSpaces(char[] s){
    for(int j = 0; j < s.length; j++){

    for(int i = s.length-3; i > j; i--){
        s[i+2] = s[i];
    }
    return s;
    }
}

I know this is O(n^2), but I don't quite understand the math behind transforming this:

(n -1) + (n - 2) + (n - 3) ... (n - n)

into this

O(n^2)

In my mind if n = 5 (String is of length 5), I would have...

(5-1) + (5-2) + (5-3) + (5-4) + (5-5) = 5(5 - ???)

which is

(n-1) + (n-2) + (n-3) + (n-4) + (n-5) = n(n - ???)

so that gives me 5*5 = 25 which is n^2. but what is the ??? I don't know what to put for the variables in the formula. I don't even know if I'm even going by this the correct way. AKA I forgot how to do math :(

Answer 1

(n -1) + (n - 2) + (n - 3) ... (n - n)

Just rewrite the following as:

1 + 2 + 3 + ....+ (n-1)

which is equal to: (n(n+1)/2 - n) .

Now you can see it is O(n^2) .

As noted by @hvd you may want to put the return statement outside the loop.

Answer 2

The Big-O notation is not the exact upper bound. It's an asymptotic upper bound. In many cases where a algorithm may look like O(n^2), but amortized analysis may show a linear order complexity.