Reverse of the sequence with recursion

Question

Hi! This is my code to solve the assignment. I used dynamic data structure "l[]". But it must bu solved without using any type of it. How can it be done?

def reverse(l):
if len(l) > 1:
print(l[len(l)-1])
del l[len(l)-1]
return reverse(l)
else:
return l[0]

def ss(n):
if n != 0:
l.append(n)
return ss(int(input()))
elif n == 0 and len(l) == 0:
return n
elif n == 0 and len(l) != 0:
l.append(n)
return reverse(l)

l = []
print(ss(int(input())))

Answer 1

So, if I get it right, what you want is to print your input in backwards order as long as input isn't 0.
Not sure how python works but maybe this could work:
def ss(n):
if n != 0:
ss(int(input()))
print(n)
else:
return

Each call will stop at another ss() call and will resume in backwards order when last call returns (when input is 0)

Answer 2

https://code.sololearn.com/ckC57fzkM61s/?ref=app

Answer 3

BlazingMagpie, thank you for working example. But it need's to print null too like:

def print_reversed(seq, pos=0):
elem = seq[pos]
if elem != 0:
print_reversed(seq, pos + 1)
print(elem)

print_reversed(tuple(map(int, sys.stdin.read().split())))

Answer 4

Thank you, guys! I solved the issue. So I need to output the null too. And here it is. Thanks for your time & help!

def reverse(n):
if n != 0:
reverse(int(input()))
print(n)
else:
print(n)
reverse(int(input()))