Question

I’m having problems reversing a list using recursion using only these functions:

def head(xs):
    return xs[0]

def tail(xs):
    return xs[1:]

def empty(xs):
    return len(xs) == 0

I can do this:

def p(xs1, xs2):
    if not empty(tail(xs1)):
        p(tail(xs1), xs2)
    xs2.append(head(xs1))

def p05(xs):
    s = []
    p(xs, s)
    return s

Is there a way to do it without using append()??

Answer 1

You could avoid changing the list in place and instead return a new list:

def p(xs1, xs2):
    if not empty(tail(xs1)):
        xs2 = p(tail(xs1), xs2)
    return xs2 + [head(xs1)]

def p05(xs):
    return p(xs, [])

You should probably change head() to return a list as well:

def head(xs):
    return xs[:1]

def tail(xs):
    return xs[1:]

and ’empty’ is not needed; [] is considered False in a python context. Then p() becomes:

def p(xs1, xs2):
    if tail(xs1):
        xs2 = p(tail(xs1), xs2)
    return xs2 + head(xs1)

Demonstration:

>>> p(range(5), [])
[4, 3, 2, 1, 0]