如何修复递归倒计时python函数的代码，使其只打印一次“LIFT OFF！”？

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

如果您只希望lift off打印一次，请打印一次：

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!')

输出：

5
4
3
2
1
LIFT OFF!

现在countdown()函数只做一件事，那就是在延迟后打印数字

Answer 2

这应该起作用：

#!/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)

以下是输出：

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