Iterative:
for (ii=1; n > 0; n--)
ii = ii * n;
return(ii);
Could be to get rid of the (1 * n) iteration when ii = 1:
for (ii=1; n > 1; n--)
ii = ii * n;
return(ii);
1
Anonimo
9 mar 2017
def factorial(n): if n = 1 return 1 else return n * factorial (n-1);
and explained how to do it without recursion.