Python return n 个列表递归出现，没有切片/深度复制或内置函数

Question

您不能使用内置函数或切片或深层复制或复制，或 while 或 for

假设这是我们的函数：

def list_occur(n):

它将 n 作为 int 参数输入和 returns 一个大小为 n 的列表，显示该系列中的元素 n（输出）：

n = 0: returns []
n = 1: returns list with size 1 that takes in index=0 [], means returns [ [] ]
n = 2: returns list with size 2 that takes in index=0 [] and index = 1 [ [] ], means returns:
[ [], [ [] ] ]
n = 3: returns list with size 3 that takes in index=0 [], index =1 [ [] ], index = 2 [ [], [ [] ] ] 
means its returns [ [], [ [] ], [ [], [ [] ] ] ]

我尝试解决这个问题，但是当我尝试 n = 2 + :

时出现错误“AttributeError: 'NoneType' object has no attribute 'append'

  def occur(n) :

    return occr_helper(n)



def occr_helper(n) :

    if (n == 0):
        return []

    return occr_helper(n-1).append(occr_helper(n-1))

    if __name__ == "__main__":

    print(occur(1))

Answer 1

为此函数 n+1 returns n 附加到 n。这可以很容易地递归编码：

num = 3

def occr(n):
    if n == 0:
        return []
    else:
        new_l = occr(n-1)
        new_l.append(occr(n-1))
        return new_l

print(occr(num))

Python return n 个列表递归出现，没有切片/深度复制或内置函数

Python return occurrence of n lists recursively without slicing / deepcopy or inbuilt functions

python

recursion

nested

list