如何修复我的递归倒计时 python 函数的代码，以便它只打印“LIFT OFF！” 一次？

Question

#!/usr/bin/env python

import time

def countdown(num):
  if num <= 0:
    return num
  else:
    time.sleep(0.1)
    print(num)
    countdown(num - 1)
    print(“LIFT OFF!”)

Answer 1

This should work:

#!/usr/bin/env python

import time

def countdown(num):
  if num <= 0:
    print('LIFT OFF!')
    return num
  else:
    time.sleep(0.1)
    print(num)
    countdown(num - 1)

here's the output:

>>> countdown(4)
4
3
2
1
LIFT OFF!
>>> 

Answer 2

If you only want lift off to print once, then print it once:

import time

def countdown(num):
  if num <= 0:
    return num
  else:
    time.sleep(0.1)
    print(num)
    countdown(num - 1)

countdown(5)
print('LIFT OFF!')

Output:

5
4
3
2
1
LIFT OFF!

Now the countdown() function does just one thing, which is print numbers after a delay.