> Figure out how much memory fib_dp is holding on to right before it returns
> the answer. fib(40) is a _big_ number! And so is fib(39), and
> fib(38), and fib(37), etc. By the time you're done, you're holding
> on to quite a huge pile of storage here. Depending on how much phys
Windson Yang writes:
> 'I'm just running them in succession and seeing how long they'. The full
> code looks like this, this is only an example.py here. and I run 'time
> python3 example.py' for each function.
>
> def fib_dp(n):
> dp = [0] * (n+1)
> if n <= 1:
> r
'I'm just running them in succession and seeing how long they'. The full
code looks like this, this is only an example.py here. and I run 'time
python3 example.py' for each function.
def fib_dp(n):
dp = [0] * (n+1)
if n <= 1:
return n
dp[0], dp[1] = 0, 1
On Sun, Aug 25, 2019 at 1:43 PM Windson Yang wrote:
>
> Thank you, Chris. I tried your suggestions. I don't think that is the reason,
> fib_dp_look() and fib_dp_set() which also allocation a big list can return in
> 2s.
(Please don't top-post)
Are you running each function more than once, or j
Thank you, Chris. I tried your suggestions. I don't think that is the
reason, fib_dp_look() and fib_dp_set() which also allocation a big list can
return in 2s.
Chris Angelico 于2019年8月25日周日 上午11:27写道：
> On Sun, Aug 25, 2019 at 12:56 PM Windson Yang wrote:
> >
> > I have two functions to calculat
On Sun, Aug 25, 2019 at 12:56 PM Windson Yang wrote:
>
> I have two functions to calculate Fibonacci numbers. fib_dp use a list to
> store the calculated number. fib_dp2 just use two variables.
>
> def fib_dp(n):
> if n <= 1:
> return n
> dp = [0] * (n+1)
>
I have two functions to calculate Fibonacci numbers. fib_dp use a list to
store the calculated number. fib_dp2 just use two variables.
def fib_dp(n):
if n <= 1:
return n
dp = [0] * (n+1)
dp[0], dp[1] = 0, 1
for i in range(2, n+1):
dp[i] =
