Develop Reference

Why does my code to reverse a Linked List only return the first node?

I am trying to reverse a linked list with the following code. I have traced it out on paper and it seems to work, but it is failing most tests- can someone let me know where it is going wrong please?
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def reverseLinkedList(head):
curr,prev,nex = head,None,head.next
while nex:
curr.next = prev
prev = curr
curr = nex
nex = nex.next
return curr

When your while loop exits, curr.next is still None, therefore it is returning only the head of the reversed linked-list.
Just add a line curr.next = prev after the while loop, then it will work fine.
So the code of the reverseLinkedList function will be:
def reverseLinkedList(head):
curr,prev,nex = head,None,head.next
while nex:
curr.next = prev
prev = curr
curr = nex
nex = nex.next
curr.next = prev
return curr
Example for better explaination:
Let's assume our linked-list is: 1 --> 2 --> 3
After execution of code linked-list becomes: 1 <-- 2 3 <-- curr
There is no link between 2 and 3, but your prev is pointing to 2.
so executing a line curr.next = prev will make the linked list:
1 <-- 2 <--3 <--curr

You can use Python's multi-variable assignments to implement the reversal (you just have to be careful with the order of variables):
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def __repr__(self):
return str(self.value) + (self.next and f"-->{self.next}" or "")
def reverseLinkedList(head):
newHead = None
while head:
newHead,newHead.next,head = head,newHead,head.next
return newHead
Output:
h = t = LinkedList(1)
for v in range(2,10):
t.next = LinkedList(v)
t = t.next
print(h)
1-->2-->3-->4-->5-->6-->7-->8-->9
rh = reverseLinkedList(h)
print(rh)
9-->8-->7-->6-->5-->4-->3-->2-->1

Select even integers in the linked list and make their sub parts

We were tasked to do the following code.
You are given a linked list that contains N integers. You are to perform the following reverse operation on the list:
Select all the subparts of the list that contain only even integers. For example, if the list is {1,2,8,9,12,16}, then the selected subparts will be {2,8}, {12,16}.
Reverse the selected subpart such as {8,2} and {16,12}.
The list should now be {1,8,2,9,16,12}.
The problem is that it is not creating the sublist and give me output like [2, 8, 12, 16], but I want it like {2,8}, {12,16}.
Below is my code:
class Node:
def __init__(self, data=None):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
self.tail = None
def append_value(self, x):
if not isinstance(x, Node):
x = Node(x)
if self.head is None:
self.head = x
else:
self.tail.next = x
self.tail = x
def reverse_list_recursive(self, current, previous):
if self.head is None:
return
elif current.next is None:
self.tail = self.head
current.next = previous
self.head = current
else:
next = current.next
current.next = previous
self.reverse_list_recursive(next, current)
def is_empty(self):
return self.head is None
def __str__(self):
to_print = ''
current = self.head
while current:
to_print += f'{current.data}->'
current = current.next
if to_print:
return f'[{to_print[:-2]}]'
return '[]'
This is my function:
def reverse_sub_parts(self):
sublist_list = list()
current = self.head
while current:
if current.data % 2 == 0:
sublist_list.append(current.data)
current = current.next
print(sublist_list)
my_list = LinkedList()
my_list.append_value(1)
my_list.append_value(2)
my_list.append_value(8)
my_list.append_value(9)
my_list.append_value(12)
my_list.append_value(16)
my_list.reverse_sub_parts()

The loop is explicitly adding all nodes that have even numbers to a single list (sublist_list). The statement sublist_list = list() is only executed once, so there is no way you could ever hope to get more than one list like that.
Secondly, even if you manage to create a variable number of lists like that, you would still need to have logic to reverse those new lists (trivial), and to replace the original sublists inside the original list with the reversed lists (less trivial).
In the end, it is not making things easier like this.
I would suggest to first make your reverse_list_recursive function more versatile, so it can also do the job for a subsection of the list, by letting it accept two more arguments: the node the precedes that section, and the node the follows it. Both would be None by default, which means that the whole list should be reversed.
I would also suggest to not use a recursive solution, as this will use stack space that is linear with the number of nodes in the list, and so if the list would have like thousands of nodes, you will run into a stack size limit.
So here is an iterative reverse function, that takes these extra, optional arguments:
def reverse(self, before_first=None, after_last=None):
prev = after_last
current = before_first.next if before_first else self.head
while current and current != after_last:
prev, current.next, current = current, prev, current.next
if before_first:
before_first.next = prev
else:
self.head = prev
Now the hardest part has been done really, as you now only need to identify the nodes that bound the sequences of even-numbered nodes:
def reverse_sub_parts(self):
prev = None
current = self.head
while current:
if current.data % 2 == 0:
if not prev or prev.data % 2 == 1:
before_start = prev
elif prev and prev.data % 2 == 0:
self.reverse(before_start, current)
prev = current
current = current.next
if prev and prev.data % 2 == 0:
self.reverse(before_start)

Deleting items in a specified index range for linked list

In my linked list, I wrote a function drop_between which deletes nodes in a specified range.
everything other than drop_between function works fine.
My expected output:
sample = Deque()
sample.push_front(3)
sample.push_front(2)
sample.push_front(1)
sample.push_front(0)
linked list = [0, 1, 2, 3]
sample.drop_between(1,3)
linked list = [0, 3] # 1, 2 removed
However, my function removes everything that is specified in the drop_between except the very first number that is passed in (which is the start).
My linked list function file:
class Node:
"""
Initialize empty node
"""
def __init__(self, data, prev = None, next = None):
self.data = data
self.next = next
self.prev = prev
class Deque:
"""
A double-ended queue
"""
def __init__(self):
"""
Initializes an empty Deque
"""
self.head = None
self.tail = None
self.size = 0
def __len__(self):
"""
Computes the number of elements in the Deque
:return: The size of the Deque
"""
return self.size
def push_front(self, e):
"""
Inserts an element at the front of the Deque
:param e: An element to insert
"""
new_head = Node(data = e, next = self.head)
if len(self) == 0:
self.tail = new_head
if self.head:
self.head.prev = new_head
self.head = new_head
self.size += 1
def drop_between(self, start, end): #need help with this function
"""
Deletes elements from the Deque that within the range [start, end)
:param start: indicates the first position of the range
:param end: indicates the last position of the range(does not drop this element)
"""
cur_index = 0
if (start > end):
raise IndexError
if (start < 0):
raise IndexError
curr = self.head
while curr.next:
last_node = curr
curr = curr.next
if ((cur_index >= start) and (cur_index < end)):
print("here", cur_index)
last_node.next = curr.next.next
curr.next.next.prev = last_node.next
cur_index += 1
def listprint(self, node):
"""
Prints each element of the node front to back
:param node:
"""
while (node is not None):
print(node.data)
last = node
node = node.next
My main file:
def main():
D = Deque()
D.push_front(9)
D.push_front(8)
D.push_front(7)
D.push_front(6)
D.push_front(5)
D.push_front(4)
D.push_front(3)
D.push_front(2)
D.push_front(1)
D.push_front(0)
D.drop_between(4,7)
D.listprint(D.head)
The output of my main displays the numbers 0-4 and 7-9 inclusive, but the output of 4 was unexpected. How am I able to properly unlink the 4 in my linked list?

Please, note that collections already have a implementation of deque
With that in mind, your task ends up really simple:
from collections import deque
def drop_between(d, start, end):
d.rotate(-start)
for __ in range(end-start):
d.popleft()
d.rotate(+start)
d = deque(n for n in range(20))
print(d)
drop_between(d, 5, 9)
print(d)
However, if you are not allowed to use this external source, I advise to code a similar logic.

Replace your drop_between function with this:
def drop_between(self, start, end): #need help with this function
"""
Deletes elements from the Deque that within the range [start, end)
:param start: indicates the first position of the range
:param end: indicates the last position of the range(does not drop this element)
"""
cur_index = 0
if (start > end):
raise IndexError
if (start < 0):
raise IndexError
#assumption: start node != head
prev_to_start_node = self.get_node_by_index(start-1)
end_node = self.get_node_by_index(end)
prev_to_start_node.next = end_node
end_node.prev = prev_to_start_node
def get_node_by_index(self,index):
i = 0
curr = self.head
while i < index and curr.next:
curr = curr.next
i+=1
return curr
Notice that I added get_node_by_index, so if you wish to remove it, you'll need to replace the code accordingly.

Python Reverse a Linked List

I am doing a Python program that implements linked list to support a few functions, one of the functions I need to do is to reverse a stack. I have made a Node, LinkedList and Stack classes, here is my Code so far:
class ListNode:
def __init__(self, Object):
self.Object = Object
self.next = None
class LinkedList:
def __init__(self):
self.head = None # first node
self.tail = None # last node
def addLast(self, Object):
newNode = ListNode(Object)
if self.head == None:
self.head = newNode
self.tail = newNode
else:
self.tail.next = newNode
self.tail = newNode
def removeFirst(self):
if self.head == None:
return
self.head = self.head.next
if self.head == None:
self.tail = None
def removeLast(self, Object):
if self.head == None:
return
current = self.head
prev = None
while current.next != None:
prev = current
current = current.next
if prev == None:
self.head = None
self.tail = None
else:
prev.next = None
self.tail = prev
def get(self, index):
current = self.head
i = 0
while i < index and current != None:
current = current.next
i = i + 1
if current != None and index >= 0:
return current.Object
else:
return None
def size(self):
current = self.head
count = 0
while current != None:
count = count + 1
current = current.next
return count
def isEmpty(self):
return self.head == None
def printList(self):
if self.head != None:
current = self.head
while current != None:
print(current.Object, end = ' ')
current = current.next
print()
# -------------------- STACK ---------------------------------------------------
class Stack:
# constructor implementation
def __init__(self):
self.llist = LinkedList()
def front(self):
return self.llist.get(0)
def dequeue(self):
self.llist.removeFirst()
def queue(self, Object):
self.llist(Object)
def push(self, Object):
self.llist.addLast(Object)
def pop(self, Object):
self.llist.removeLast(Object)
def printStack(self):
self.llist.printList()
def size(self):
return self.llist.size()
def isEmpty(self):
return self.llist.isEmpty()
# ----------------------- Reverse LIST ------------------------------
def Reverse(S):
# S is a Stack Object
Here is my attempt at the problem:
def rRecursive( self ) :
self._rRecursive( self.head )
def _reverseRecursive( self, n ) :
if None != n:
right = n.next
if self.head != n:
n.next = self.head
self.head = n
else:
n.next = None
self._rRecursive( right )
def Reverse(S):
s.rRecursive()
If this was an ordinary list I could easily reverse it using [::-1] but I cannot since the linkedlist is an object. I was thinking maybe I could use temporary values so I can someone append the beginning of the stack to a list and then somehow convert it back into a stack object.
Edit for duplication: The goal of my program is use an existing Stack and reverse it. The linked post deals with a linked list that was already in a list format.
def get(self, index):
current = self.head
i = 0
while i < index and current != None:
current = current.next
i = i + 1
if current != None and index >= 0:
return current.Object
else:
return None
Edit 2: added my get function.
class Stack:
# constructor implementation
def __init__(self):
self.llist = LinkedList()
def front(self):
return self.llist.get(0)
# push method implementation
def push(self, Object):
self.llist.addLast(Object)
def pop1(self):
self.llist.removeLast1()
def Reverse(S):
new_stack = Stack()
while not S.isEmpty():
new_stack.push(S.front())
S.pop1()
return new_stack
# Current Stack after Push: 12 14 40 13
# Stack after Pop: 12 14 40
# Stack after Reversal: 12 12 12
Edit 3: Added a rework of my code, it returns back a wrong reversal with the first element over and over again.

It's quite easy to reverse a stack using basic stack operations. Pop each item off the old stack and push it onto a new one.
def reverse(stack):
new_stack = Stack()
while not stack.isEmpty():
new_stack.push(stack.front())
stack.pop()
return new_stack
You could do something similar to destructively reverse the LinkedList within the Stack while reusing the stack object itself, you'd just need to use the list operations rather than the stack operations that are aliased to them.
Speaking of stack operations, you'll probably find that your stack performs better if you push and pop from the front, rather than the back. Removing an item from the end of the linked list requires iterating over the whole list (to find the next-to-last node). In contrast, both adding and removing from the front are fast.

You shouldn't have to fiddle with link pointers in your reversal function. I assume that you have a pop() method and other basics with your stack; if not, then clone your removeFirst function to return the removed node.
Now, the recursive function is simple: pop the head of the list, reverse the remaining stack (if any), and add the popped node to the end. Does this handle your problem?
def reverseStack(self):
move_me = self.pop()
if not self.isEmpty():
return (self.reverseStack()).addLast(move_me)
else:
new_stack = Stack()
return new_stack.addLast(move_me)

Other approaches to solving this problem: Cheat.
Define a __iter__ method (which makes sense in any event; iteration is a core Python behavior) to make your type iterable. Simple example:
def __iter__(self):
cur = self.head
while cur is not None:
yield cur.Object
cur = cur.next
Then just do:
values = list(self) # Creates a list containing the current set of values
self.head = self.tail = None # Clear existing linked list
# Add back all the values in reverse order
for value in reversed(values):
self.addLast(value)
Sure, it's likely less efficient in memory allocation/deallocation overhead than doing it properly. But the effect is likely marginal, and it simplifies the implementation code dramatically.
Of course, doing it properly isn't that hard, it's just slightly more confusing (totally untested, but it should be close to right):
def reverse(self):
# Start at beginning, which will be new end
cur, last = self.head, None
# Reverse head and tail pointers in advance
self.head, self.tail = self.tail, self.head
# Traverse while reversing direction of each node pointer
while cur is not None:
# Tuple pack and unpack allows one-line variable swap
cur.next, cur, last = last, cur.next, cur

The usual algorithm for reversing things using basic data structures is to use the fact that a stack is a first in last out data structure. That means if you pop until the stack is empty, you will get the items in the opposite order you pushed them on.
But it looks like you want to do this via recursive functions rather than an explicit stack - in this case, your function would normally fetch the front element,then recurse on the rest of the data structure, then handle that first element. This gets all the elements off in order, but handles them in the opposite order as each recursive call finishes and you work your way back up the call stack. If you neex to add the elements to a reversed data structure (and not just print them), you can build it up via return values by noticing that once you have the reverse if everything after the current element, you just need to attach that element to the front and you have the reverse of everything you were handed.

class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
def Reverse(head):
if head == None :
return None
elif head.next == None :
return head # returns last element from stack
else :
temp = head.next # stores next element while moving forward
head.next = None # removes the forward link
li = Reverse(temp) # returns the last element
temp.next = head # now do the reverse link here
return li
def main() :
temp4 = Node(4,None)
temp3 = Node(3,temp4)
temp2 = Node(2,temp3)
head = Node(1,temp2)
res = Reverse(head)
while res != None :
print(res.data)
res = res.next
main()

Reversing a linked list in python

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 implement the function reverse_linked_list in this way:
def reverse_linked_list(head):
temp = head
head = None
temp1 = temp.next
temp2 = temp1.next
temp1.next = None
temp2.next = temp1
temp1.next = temp
return temp2
class Node(object):
def __init__(self,value=None):
self.value = value
self.next = None
def to_linked_list(plist):
head = None
prev = None
for element in plist:
node = Node(element)
if not head:
head = node
else:
prev.next = node
prev = node
return head
def from_linked_list(head):
result = []
counter = 0
while head and counter < 100: # tests don't use more than 100 nodes, so bail if you loop 100 times.
result.append(head.value)
head = head.next
counter += 1
return result
def check_reversal(input):
head = to_linked_list(input)
result = reverse_linked_list(head)
assert list(reversed(input)) == from_linked_list(result)
It is called in this way: check_reversal([1,2,3]). The function I have written for reversing the list is giving [3,2,1,2,1,2,1,2,1] and works only for a list of length 3. How can I generalize it for a list of length n?

The accepted answer doesn't make any sense to me, since it refers to a bunch of stuff that doesn't seem to exist (number, node, len as a number rather than a function). Since the homework assignment this was for is probably long past, I'll post what I think is the most effective code.
This is for doing a destructive reversal, where you modify the existing list nodes:
def reverse_list(head):
new_head = None
while head:
head.next, head, new_head = new_head, head.next, head # look Ma, no temp vars!
return new_head
A less fancy implementation of the function would use one temporary variable and several assignment statements, which may be a bit easier to understand:
def reverse_list(head):
new_head = None # this is where we build the reversed list (reusing the existing nodes)
while head:
temp = head # temp is a reference to a node we're moving from one list to the other
head = temp.next # the first two assignments pop the node off the front of the list
temp.next = new_head # the next two make it the new head of the reversed list
new_head = temp
return new_head
An alternative design would be to create an entirely new list without changing the old one. This would be more appropriate if you want to treat the list nodes as immutable objects:
class Node(object):
def __init__(self, value, next=None): # if we're considering Nodes to be immutable
self.value = value # we need to set all their attributes up
self.next = next # front, since we can't change them later
def reverse_list_nondestructive(head):
new_head = None
while head:
new_head = Node(head.value, new_head)
head = head.next
return new_head

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

Here is a way to reverse the list 'in place'. This runs in constant time O(n) and uses zero additional space.
def reverse(head):
if not head:
return head
h = head
q = None
p = h.next
while (p):
h.next = q
q = h
h = p
p = h.next
h.next = q
return h
Here's an animation to show the algorithm running.
(# symbolizes Null/None for purposes of animation)

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.

You can do the following to reverse a singly linked list (I assume your list is singly connected with each other).
First you make a class Node, and initiate a default constructor that will take the value of data in it.
class Node:
def __init__(self, data):
self.data = data
self.next = None
This solution will reverse your linked list "iteratively".
I am making a class called SinglyLinkedList which will have a constructor:
class SinglyLinkedList:
def __init__(self):
self.head = None
then I have written a method to reverse the list, print the length of the list, and to print the list itself:
# method to REVERSE THE LINKED LIST
def reverse_list_iterative(self):
prev = None
current = self.head
following = current.next
while (current):
current.next = prev
prev = current
current = following
if following:
following = following.next
self.head = prev
# Method to return the length of the list
def listLength(self):
count = 0
temp = self.head
while (temp != None):
temp = temp.next
count += 1
return count
# Method to print the list
def printList(self):
if self.head == None:
print("The list is empty")
else:
current_node = self.head
while current_node:
print(current_node.data, end = " -> ")
current_node = current_node.next
if current_node == None:
print("End")`
After that I hard code the list, and its contents and then I link them
if __name__ == '__main__':
sll = SinglyLinkedList()
sll.head = Node(1)
second = Node(2)
third = Node(3)
fourth = Node(4)
fifth = Node(5)
# Now linking the SLL
sll.head.next = second
second.next = third
third.next = fourth
fourth.next = fifth
print("Length of the Singly Linked List is: ", sll.listLength())
print()
print("Linked List before reversal")
sll.printList()
print()
print()
sll.reverse_list_iterative()
print("Linked List after reversal")
sll.printList()
Output will be:
Length of the Singly Linked List is: 5
Linked List before reversal 1 -> 2 -> 3 -> 4 -> 5 -> End
Linked List after reversal
5 -> 4 -> 3 -> 2 -> 1 -> End

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.

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)

def reverseLinkedList(head):
current = head
previous = None
nextNode = None
while current:
nextNode = current.nextNode
current.nextNode = previous
previous = current
current = nextNode
return previous

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-

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

class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList :
def __init__(self):
self.head = None
def add(self, data):
node = Node(data)
if not self.head:
self.head = node
else:
current = self.head
while current.next != None:
current = current.next
current.next = node
def printList(self):
value = []
if not self.head:
print("liss is Empty")
else:
current = self.head
while current:
value.append(current.data)
current = current.next
print(value)
# this func start reverse list from the last to first
def reverseLinkedList(self,node1,node2):
if self.head == None:
print("list Empty")
else:
# when we reach the last of list we link head with the last element and we disconnect head with second element that will make first element in the last of the list
if node2 == None and node1 != None:
self.head.next = None
self.head = node1
return
else:
self.reverseLinkedList(node1.next, node2.next )
node2.next = node1
ln = LinkedList()
ln.add(1)
ln.add(2)
ln.add(3)
ln.add(4)
ln.add(5)
ln.add(6)
ln.add(7)
ln.add(8)
ln.add(9)
ln.printList()
ln.reverseLinkedList(ln.head,ln.head.next)
print("after first reverse")
ln.printList()
# after i reverse list I add new item to the last
ln.add(0)
print("after add new element to the last of the list")
ln.printList()
print("after second reverse")
# i made second reverse to check after I add new element if my function work perfectly
ln.reverseLinkedList(ln.head,ln.head.next)
ln.printList()
Output :
[1, 2, 3, 4, 5, 6, 7, 8, 9]
after first reverse
[9, 8, 7, 6, 5, 4, 3, 2, 1]
after add new element to the last of the list
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
after second reverse
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Efficient way to reverse linkedlist is discussed in the below steps. Approach to reverse the linking.
Use a variable curr and equal it to head.
Use a variable prev and equal it to None.
Apply the loop with the condition that loop will run till curr is not None and in this condition use another variable next which is equal to curr.next. This will be used to get hold of next node. Now make curr.next = prev. In this way the head after reversing linkedlist will point to None. Now make prev = curr and curr = next
Code snippet
def reverse_linked_list(head):
# corner case
if head == None:
return
curr = head
# reason being orginal head after reversing should point to None
prev = None
while curr is not None:
next = curr.next
curr.next = prev
prev = curr
curr = next
# prev is returned because both curr and next will be None after reversing
return prev
There is another way to perform the task which will take time complexity as theta(n) and space complexity as theta(n). This is a recursive solution and steps are as below
Step 1: Apply the recursion for n -1 node from the end side. This means reverse the list leaving the first node.
Step 2: Link the first node with all nodes obtained from step 1 in reverse order
Code
def reverse_linked_list_recursion(head):
# corner case
if head == None:
return
if head.next == None:
return head
rest_head = reverse_linked_list_recursion(head.next)
rest_tail = head.next
rest_tail.next = head
head.next = None
return rest_head

U can use mod function to get the remainder for each iteration and obviously it will help reversing the list . I think you are a student from Mission R and D
head=None
prev=None
for i in range(len):
node=Node(number%10)
if not head:
head=node
else:
prev.next=node
prev=node
number=number/10
return head