How do I reverse a list using recursion in Python?

问题:

I want to have a function that will return the reverse of a list that it is given -- using recursion. How can I do that?

回答1:

Append the first element of the list to a reversed sublist:

mylist = [1, 2, 3, 4, 5]
backwards = lambda l: (backwards (l[1:]) + l[:1] if l else []) 
print backwards (mylist)

回答2:

A bit more explicit:

def rev(l):
    if len(l) == 0: return []
    return [l[-1]] + rev(l[:-1])

---

This turns into:

def rev(l):
    if not l: return []
    return [l[-1]] + rev(l[:-1])

Which turns into:

def rev(l):
    return [l[-1]] + rev(l[:-1]) if l else []

Which is the same as another answer.

---

Tail recursive / CPS style (which python doesn't optimize for anyway):

def rev(l, k):
    if len(l) == 0: return k([])
    def b(res):
        return k([l[-1]] + res)
    return rev(l[:-1],b)


>>> rev([1, 2, 3, 4, 5], lambda x: x)
[5, 4, 3, 2, 1]

回答3:

I know it's not a helpful answer (though this question has been already answered), but in any real code, please don't do that. Python cannot optimize tail-calls, has slow function calls and has a fixed recursion depth, so there are at least 3 reasons why to do it iteratively instead.

回答4:

The trick is to join after recursing:

def backwards(l):
  if not l:
    return
  x, y = l[0], l[1:]
  return backwards(y) + [x]

回答5:

This one reverses in place. (Of course an iterative version would be better, but it has to be recursive, hasn't it?)

def reverse(l, first=0, last=-1):
    if first >= len(l)/2: return
    l[first], l[last] = l[last], l[first]
    reverse(l, first+1, last-1)

mylist = [1,2,3,4,5]
print mylist
reverse(mylist)
print mylist

回答6:

def reverse(q):
    if len(q) != 0:
        temp = q.pop(0)
        reverse(q)
        q.append(temp)
    return q

回答7:

def revList(alist):
    if len(alist) == 1:       
        return alist #base case
    else:
        return revList(alist[1:]) + [alist[0]]

print revList([1,2,3,4])
#prints [4,3,2,1]

回答8:

looks simpler:

    def reverse (n):
        if not n: return []
        return [n.pop()]+reverse(n)

回答9:

Use the Divide & conquer strategy. D&C algorithms are recursive algorithms. To solve this problem using D&C, there are two steps:

1. Figure out the base case. This should be the simplest possible case.
2. Divide or decrease your problem until it becomes the base case.

Step 1: Figure out the base case. What’s the simplest list you could get? If you get an list with 0 or 1 element, that’s pretty easy to sum up.

if len(l) == 0:  #base case
    return []

Step 2: You need to move closer to an empty list with every recursive call

recursive(l)    #recursion case

for example

l = [1,2,4,6]
def recursive(l):
    if len(l) == 0:
        return []  # base case
    else:
        return [l.pop()] + recursive(l)  # recusrive case


print recursive(l)

>[6,4,2,1]

Source : Grokking Algorithms

回答10:

Take the first element, reverse the rest of the list recursively, and append the first element at the end of the list.

回答11:

def reverseList(listName,newList = None):
if newList == None:
    newList = []
if len(listName)>0:
    newList.append((listName.pop()))
    return reverseList(listName, newList)
else:
    return newList

print reverseList([1,2,3,4]) [4,3,2,1]

回答12:

Using Mutable default argument and recursion :

def hello(x,d=[]):
    d.append(x[-1])
    if len(x)<=1:
        s="".join(d)
        print(s)

    else:
        return hello(x[:-1])

hello("word")

additional info

x[-1]    # last item in the array
x[-2:]   # last two items in the array
x[:-2]   # everything except the last two items

Recursion part is hello(x[:-1]) where its calling hello function again after x[:-1]

回答13:

Why not:

a = [1,2,3,4,5]
a = [a[i] for i in xrange(len(a)-1, -1, -1)] # now a is reversed!

回答14:

This will reverse a nested lists also!

A = [1, 2, [31, 32], 4, [51, [521, [12, 25, [4, 78, 45], 456, [444, 111]],522], 53], 6]

def reverseList(L):

    # Empty list
    if len(L) == 0:
        return

    # List with one element
    if len(L) == 1:

        # Check if that's a list
        if isinstance(L[0], list):
            return [reverseList(L[0])]
        else:
            return L

    # List has more elements
    else:
        # Get the reversed version of first list as well as the first element
        return reverseList(L[1:]) + reverseList(L[:1])

print A
print reverseList(A)

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