递归函数返回元组python

Question

I want to build a RECURSIVE function that takes a natural number n and returns a tuple of numbers starting with 0 and ending before n. So rec_range(5) returns (0,1,2,3,4) and rec_range(1) returns (0, ).

This is what I have so far:

def rec_range(n):
"""returns a tuple of numbers starting with 0 and ending before n

natural number -> tuple of numbers"""
   if n = 0:
       return 0
   else:
       return rec_range(0, n)

I'm not sure what to do next. Also, it should be noticed that I could not test this function because there was an invalid syntax error.

Answer 1

In python you can concatenate tuples with + .

def rec_range(n):
    if n == 0:
        return (0,)
    else:
        return rec_range(n - 1) + (n,)

Answer 2

Define rec_range(n) as returning a tuple of counts < n, in increasing order. This mimics the set theory definition of the natural numbers. Unless an instructor mandates differently, rec_range(0) should (logically) be an empty tuple ().

Repeated concatention of tuples turns what should be an O(n) function (appending to a list) into an O(n*n) function. If a tuple rather than list is required, the conversion to tuple can be the last step. Here is an O(n) tail_recursive solution.

def rec_range(n, answer=None):
    if answer is None:
        answer = []
    if n > 0:
        n -= 1
        answer.append(n)
        return rec_range(n, answer)
    else:
        return tuple(reversed(answer))

print(rec_range(0), rec_range(1), rec_range(10))
#() (0,) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)