Why is the time complexity O(n^2) in this code?

问题:

I just didn't get it, why the time complexity is O(n^2) instead of O(n*logn)? The second loop is incrementing 2 each time so isn't it O(logn)?

void f3(int n){
  int i,j,s=100;
  int* ar = (int*)malloc(s*sizeof(int));

  for(i=0; i<n; i++){
    s=0;
    for(j=0; j<n; j+=2){
      s+=j;
      printf("%d\n", s);
  }
  free(ar);
}

回答1:

By incrementing by two, rather than one, you're doing the following N*N*(1/2). With big(O) notation, you don't care about the constant, so it's still N*N. This is because big(O) notation reference the complexity of the growth of an algorithm.

回答2:

Outer loop will execute for n times and for each outer loop iteration inner loop will execute for n/2 times because j +=2

Order = n×n/2 = n^2/2 =O(n^2) because constant doesn't affect run time for large n

回答3:

there is an increment of 2 so loop will runs for n/2.So the complexity will be n*n/2. if increament happened in GP like 2