Reversing a linked list in python

Question

I am asked to reverse a which takes head as parameter where as head is a linked list e.g.: 1 -> 2 -> 3 which was returned from a function already defined I tried to implemen

Answer 1

Here is the whole thing in one sheet. Contains the creation of a linked list, and code to reverse it.

Includes an example so you can just copy and paste into an idle .py file and run it.

class Node(object):
    def __init__(self, value, next=None): 
        self.value = value                
        self.next = next                  


def reverse(head):
    temp = head
    llSize = 0
    while temp is not None:
        llSize += 1
        temp = temp.next
    for i in xrange(llSize-1,0,-1):
        xcount = 0
        temp = head
        while (xcount != i):
            temp.value, temp.next.value = temp.next.value, temp.value
            temp = temp.next
            xcount += 1
    return head


def printnodes(n):
    b = True
    while b == True:
        try:
            print n.value
            n = n.next
        except:
            b = False

n0 = Node(1,Node(2,Node(3,Node(4,Node(5,)))))
print 'Nodes in order...'
printnodes(n0)
print '---'
print 'Nodes reversed...'
n1 = reverse(n0)
printnodes(n1)

Answer 2

Node class part borrowed from interactive python.org: http://interactivepython.org/runestone/static/pythonds/BasicDS/ImplementinganUnorderedListLinkedLists.html

I created the reversed function. All comments in the loop of reverse meant for 1st time looping. Then it continues.

class Node():
  def __init__(self,initdata):
    self.d = initdata
    self.next = None

  def setData(self,newdata):
    self.d = newdata

  def setNext(self,newnext):
    self.next = newnext

  def getData(self):
    return self.d

  def getNext(self):
    return self.next

class LinkList():
  def __init__(self):
    self.head = None

  def reverse(self):
    current = self.head   >>> set current to head(start of node)
    previous = None       >>>  no node at previous
    while current !=None: >>> While current node is not null, loop
        nextt =  current.getNext()  >>> create a pointing var to next node(will use later)
        current.setNext(previous)   >>> current node(or head node for first time loop) is set to previous(ie NULL), now we are breaking the link of the first node to second node, this is where nextt helps(coz we have pointer to next node for looping)
        previous = current  >>> just move previous(which was pointing to NULL to current node)
        current = nextt     >>> just move current(which was pointing to head to next node)

    self.head = previous   >>> after looping is done, (move the head to not current coz current has moved to next), move the head to previous which is the last node.

Answer 3

Following is the generalized code to reverse a singly linked list, where head is given as function's argument:

def reverseSll(ll_head):
    # if head of the linked list is empty then nothing to reverse
    if not ll_head:
        return False
    # if only one node, reverse of one node list is the same node
    if not ll_head.next:
        return ll_head
    else:
        second = ll_head.next # get the second node of the list
        ll_head.next = None # detach head node from the rest of the list
        reversedLL = reverseSll(second) # reverse rest of the list
        second.next = ll_head # attach head node to last of the reversed list
        return reversedLL

Let me explain what I am doing here:

1) if head is null or head.next is null(only one node left in the list) return node
2) else part: take out 1st node, remove its link to rest of the list, reverse rest of the list(reverseSll(second)) and add 1st node again at last and return the list

Github link for the same

Answer 4

Most previous answers are correct but none of them had the complete code including the insert method before and and after the reverse so you could actually see the outputs and compare. That's why I'm responding to this question. The main part of the code of course is the reverse_list() method. This is in Python 3.7 by the way.

class Node(object):
    def __incurrent__(self, data=None, next=None):
        self.data = data
        self.next = next


class LinkedList(object):

    def __incurrent__(self, head=None):
        self.head = head

    def insert(self, data):
        tmp = self.head
        self.head = Node(data)
        self.head.next = tmp

    def reverse_list(self):
        current = self.head
        prev = None

        while current :
            #create tmp to point to next
            tmp = current.next
            # set the next to point to previous
            current.next = prev
            # set the previous to point to current
            prev = current
            #set the current to point to tmp
            current = tmp
        self.head = prev


    def print(self):
        current = self.head
        while current != None:
            print(current.data,end="-")
            current = current.next
        print(" ")


lk = LinkedList()
lk.insert("a")
lk.insert("b")
lk.insert("c")

lk.print()
lk.reverse_list()
lk.print()

output:

c-b-a- 
a-b-c- 

Answer 5

I found blckknght's answer useful and it's certainly correct, but I struggled to understand what was actually happening, due mainly to Python's syntax allowing two variables to be swapped on one line. I also found the variable names a little confusing.

In this example I use previous, current, tmp.

def reverse(head):
    current = head
    previous = None

    while current:
        tmp = current.next
        current.next = previous   # None, first time round.
        previous = current        # Used in the next iteration.
        current = tmp             # Move to next node.

    head = previous

Taking a singly linked list with 3 nodes (head = n1, tail = n3) as an example.

n1 -> n2 -> n3

Before entering the while loop for the first time, previous is initialized to None because there is no node before the head (n1).

I found it useful to imagine the variables previous, current, tmp 'moving along' the linked list, always in that order.

First iteration

previous = None

[n1] -> [n2] -> [n3] current tmp current.next = previous

Second iteration

[n1] -> [n2] -> [n3] previous current tmp current.next = previous

Third iteration

# next is None

[n1] -> [n2] -> [n3] previous current current.next = previous

Since the while loop exits when current == None the new head of the list must be set to previous which is the last node we visited.

Edited

Adding a full working example in Python (with comments and useful str representations). I'm using tmp rather than next because next is a keyword. However I happen to think it's a better name and makes the algorithm clearer.

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

    def __str__(self):
        return str(self.value)

    def set_next(self, value):
        self.next = Node(value)
        return self.next


class LinkedList:
    def __init__(self, head=None):
        self.head = head

    def __str__(self):
        values = []
        current = self.head
        while current:
            values.append(str(current))
            current = current.next

        return ' -> '.join(values)

    def reverse(self):
        previous = None
        current = self.head

        while current.next:
            # Remember `next`, we'll need it later.
            tmp = current.next
            # Reverse the direction of two items.
            current.next = previous
            # Move along the list.
            previous = current
            current = tmp

        # The loop exited ahead of the last item because it has no
        # `next` node. Fix that here.
        current.next = previous

        # Don't forget to update the `LinkedList`.
        self.head = current


if __name__ == "__main__":

    head = Node('a')
    head.set_next('b').set_next('c').set_next('d').set_next('e')

    ll = LinkedList(head)
    print(ll)
    ll.revevse()
    print(ll)

Results

a -> b -> c -> d -> e
e -> d -> c -> b -> a

Answer 6

I tried a different approach, in place reversal of the LList. Given a list 1,2,3,4

If you successively swap nearby nodes,you'll get the solution.

len=3 (size-1)
2,1,3,4
2,3,1,4
2,3,4,1

len=2 (size-2)
3,2,4,1
3,4,2,1

len=1 (size-3)
4,3,2,1

The code below does just that. Outer for loop successively reduces the len of list to swap between. While loop swaps the data elements of the Nodes.

def Reverse(head):
    temp = head
    llSize = 0
    while temp is not None:
        llSize += 1
        temp = temp.next


    for i in xrange(llSize-1,0,-1):
        xcount = 0
        temp = head
        while (xcount != i):
            temp.data, temp.next.data = temp.next.data, temp.data
            temp = temp.next
            xcount += 1
    return head

This might not be as efficient as other solutions, but helps to see the problem in a different light. Hope you find this useful.