Question

def digits(n):
    res = []
    while n > 0:
        res.append(n % 10)
        n /= 10
    return res

I want to rewrite this function so it uses recursion. I’m currently lost as to what to do. Can anyone give me some direction?

Answer 1

Here’s a possible solution:

def digits(n):
    if n < 10:
        return [n]
    return digits(n/10) + [n%10]

digits(123)
> [1, 2, 3]

The above solution fixes a bug in your code, you were returning the digits in reverse order. Also notice that n must be an integer greater than or equal to zero for producing correct results.

Here’s how it works:

1. If the number is less than 10, then return a list with the number, as there are no more digits to be processed
2. If the number is greater than 9, get the last digit in the current number and add it to the end of the list that results of recursively calling digits on a smaller number – namely, the number without the last digit that we just processed.

The call to digits(123) will look like this at each step of the recursion:

digits(123) = digits(123/10) + [3]
digits(12)  = digits(12/10)  + [2]
digits(1)   = [1]

Now we go up the call stack:

[1]
[1] + [2]
[1, 2] + [3]
[1, 2, 3]

EDIT :

Accepting @thg435’s challenge, here’s a tail-recursive solution:

def digits(n):
    def loop(i, acc):
        if i < 10:
            return [i] + acc
        return loop(i/10, [i%10] + acc)
    return loop(n, [])