找出大O时间的复杂性

Question

i = 1;     
while (i <= n) {
     j = n - i;
     while (j >= 2) {
         for (k = 1; k <= j; k++) {
             s = s + Arr[k];
         }
         j = j - 2;
     }
     i = i + 1;
}

The part that confuses me is where it says

j = n - i;
while(j >= 2){

I'm not really sure how to show my work on that part. I'm pretty sure the algorthim is O(n^3) though.

Answer 1

You can simplify it a bit in order to see things more clearly:

for(i = 1; i <= n; i++)
{
    for(j = n - i; j >= 2; j -= 2)
    {
        for(k = 1; k <= j; k++)
        {
            s = s + Arr[k];
        }
    }
}

Now things should be simpler

• for(i = 1; i <= n; i++) : O(n) [executes exactly n times, actually]
• for(j = n - i; j >= 2; j -= 2) : (n-1)/2 in 1st iteration, (n-3)/2 in the 2nd and so on... O(n)
• for(k = 1; k <= j; k++) n-2 in 1st iteration, n-3 in the 2nd and so on... O(n)
• s = s + Arr[k]; [simple operation] : O(1)

Multiply every step and you get O(n^3)

If you are still having trouble with it, I would suggest you run a few simulations of this code with varying n values and a counter inside the loops. Hopefully you'll be able to see how the O(n) is the complexity for each loop