Reverse a LinkedList (easy)

We'll cover the following

- - Problem Statement
- - Try it yourself
- - Solution
- - - Code
- - - Time complexity
    - Space complexity

Problem Statement #

Given the head of a Singly LinkedList, reverse the LinkedList. Write a function to return the new head of the reversed LinkedList.

Try it yourself #

Try solving this question here:

Java

Python3

C++

from __future__ import print_function
 
 
class Node:
  def __init__(self, value, next=None):
    self.value = value
    self.next = next
 
  def print_list(self):
    temp = self
    while temp is not None:
      print(temp.value, end=" ")
      temp = temp.next
    print()
 
 
def reverse(head):
  # TODO: Write your code here
  return head
 
 
def main():
  head = Node(2)
  head.next = Node(4)
  head.next.next = Node(6)
  head.next.next.next = Node(8)
  head.next.next.next.next = Node(10)
 
  print("Nodes of original LinkedList are: ", end='')
  head.print_list()
  result = reverse(head)
  print("Nodes of reversed LinkedList are: ", end='')
  result.print_list()
 
 
main()
 

Output

0.464s

Nodes of original LinkedList are: 2 4 6 8 10 Nodes of reversed LinkedList are: 2 4 6 8 10

Solution #

To reverse a LinkedList, we need to reverse one node at a time. We will start with a variable current which will initially point to the head of the LinkedList and a variable previous which will point to the previous node that we have processed; initially previous will point to null.

In a stepwise manner, we will reverse the current node by pointing it to the previous before moving on to the next node. Also, we will update the previous to always point to the previous node that we have processed. Here is the visual representation of our algorithm:

1 of 7

2 of 7

3 of 7

4 of 7

5 of 7

6 of 7

7 of 7

Code #

Here is what our algorithm will look like:

Java

Python3

C++

from __future__ import print_function
 
 
class Node:
  def __init__(self, value, next=None):
    self.value = value
    self.next = next
 
  def print_list(self):
    temp = self
    while temp is not None:
      print(temp.value, end=" ")
      temp = temp.next
    print()
 
 
def reverse(head):
  previous, current, next = None, head, None
  while current is not None:
    next = current.next  # temporarily store the next node
    current.next = previous  # reverse the current node
    previous = current  # before we move to the next node, point previous to the current node
    current = next  # move on the next node
  return previous
 
 
def main():
  head = Node(2)
  head.next = Node(4)
  head.next.next = Node(6)
  head.next.next.next = Node(8)
  head.next.next.next.next = Node(10)
 
  print("Nodes of original LinkedList are: ", end='')
  head.print_list()
  result = reverse(head)
  print("Nodes of reversed LinkedList are: ", end='')
  result.print_list()
 
 
main()
 

Output

0.463s

Nodes of original LinkedList are: 2 4 6 8 10 Nodes of reversed LinkedList are: 10 8 6 4 2

Time complexity #

The time complexity of our algorithm will be $O(N)$ where ‘N’ is the total number of nodes in the LinkedList.

Space complexity #

We only used constant space, therefore, the space complexity of our algorithm is $O(1)$ .

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Introduction

Reverse a Sub-list (medium)