如何从递归 function 中获取数字列表并将其作为列表返回？

Question

I made a recursive function that gives me a certain number if it's odd or even. So, if I call the function with a 5 , it'll output 5, 16, 8, 4, 2, 1 . How do I have the output return this sequence of numbers as a list? So, [5, 16, 8, 4, 2, 1] .

This is my code:

def recursion_list(number):
    ls = []
    print(number)
    if number == 1:
        return 1
    if number % 2 == 0:
        first = recursion_list(number // 2)
        return first
    elif number % 2 == 1:
        second = recursion_list(3 * number + 1)
        return second

Answer 1

There are two ways to do this.

First, return a variable of list type and receive the return value and appending it. But this method has a large overhead

Second, make variable ls to global variable of list type and instead of returning a value, attach it to the ls variable.

Answer 2

Try to return a list of the printed variables and recursively append them.

def recursion_list(number):
    ls = [number]
    print(number)
    if number == 1:
        return ls
    if number % 2 == 0:
        ls += recursion_list(number // 2)
        return ls
    elif number % 2 == 1:
        ls += recursion_list(3 * number + 1)
        return ls
print(recursion_list(5))

output

5
16
8
4
2
1
[5, 16, 8, 4, 2, 1]

Answer 3

def recursion_list(number):
   ls = [number]
   if number == 1:
      return ls
   elif number % 2 == 0:
      return ls + recursion_list(number // 2)
   elif number % 2 == 1:
      return ls + recursion_list(3 * number + 1)

Answer 4

The task can be done by defining a list globally.

ls = []
def recursion_list(number):
    ls.append(number)
    if number == 1:
        return ls
    if number % 2 == 0:
        first = recursion_list(number // 2)
        return first
    elif number % 2 == 1:
        second = recursion_list(3 * number + 1)
        return second

print(recursion_list(5))

Answer 5

You could convert you recursive function into a generator and make a list of the output only when needed.

def recursion_gen(number):
    yield number
    if number == 1: return
    if number % 2 == 0:
        for n in recursion_gen(number // 2):
            yield n
    else:
        for n in recursion_gen(3 * number + 1):
            yield n

print(*recursion_gen(5)) # 5 16 8 4 2 1

ls = list(recursion_gen(5))
print(ls) # [5, 16, 8, 4, 2, 1]

To return a list directly, you don't need an intermediate variable:

def recursion_list(number):
    if number == 1:
        return [number]
    if number % 2 == 0:
        return [number] + recursion_list(number // 2)
    else:
        return [number] + recursion_list(3 * number + 1)