如何创建一个返回从n到1的整数列表的函数？

Question

How do I write a function called countdown that counts down starting from n and goes until 1 ? The function should return a list , the contents of which should be integers going from n down to 1.

def countdown(n):
    if n >= 1:
        countdown(n-1)
    print(n)

Answer 1

Since you want to return a list , you need to create that list in the function.

def countdown(n):
    return list(range(n, 0, -1))

range creates your sequence from n to 0 (non-inclusive, which means it'll stop at 1), with a step of -1 each time.

list then converts the sequence into the list that you want returned.

This also means that you don't actually have to create a specific function for a countdown list. You can just directly call list(range(n, 0, -1)) .

Answer 2

Using recursion:

def countdown(n):
    if n < 1:
        return []
    return [n] + countdown(n-1)

This approach provides the "base case" and creation of a list of integers once the base is reached.

Check out this link to visualize the execution. Do let me know if you have any questions.