Reverse a Singly Linked List Iterative in DSA Python - Build from Scratch

Reverse a Singly Linked List Iterative

📖 Scenario: You are working on a simple program that manages a list of tasks. Each task is stored as a node in a singly linked list. Sometimes, you want to reverse the order of tasks to see them from last to first.

🎯 Goal: Build a program that creates a singly linked list with specific tasks, then reverses the list using an iterative method, and finally prints the reversed list.

📋 What You'll Learn

Create a singly linked list with nodes containing the exact values: 'Task1', 'Task2', 'Task3', 'Task4'

Use variables named head, prev, current, and next_node as needed

Implement the iterative reversal of the singly linked list

Print the reversed linked list in the format: Task4 -> Task3 -> Task2 -> Task1 -> None

💡 Why This Matters

🌍 Real World

Reversing linked lists is useful in many applications like undo operations, navigation history, and data processing pipelines where order reversal is needed.

💼 Career

Understanding linked list reversal is a common interview question and helps build foundational skills for working with dynamic data structures in software development.

1

Create the singly linked list nodes

Define a class called Node with an __init__ method that takes data and sets self.data and self.next. Then create four nodes with data 'Task1', 'Task2', 'Task3', and 'Task4'. Link them so that head points to the first node.

DSA Python

# Define Node class and create linked list nodes
# Your code here

Hint

Start by defining the Node class with data and next attributes. Then create four nodes and link them by setting the next attribute of each node to the next node.

2

Set up variables for iterative reversal

Create two variables: prev and current. Set prev to None and current to head.

DSA Python

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

head = Node('Task1')
second = Node('Task2')
third = Node('Task3')
fourth = Node('Task4')

head.next = second
second.next = third
third.next = fourth

# Create variables prev and current for reversal
# Your code here

Hint

Set prev to None because before reversal there is no previous node. Set current to head to start from the first node.

3

Iteratively reverse the linked list

Use a while loop that runs while current is not None. Inside the loop, create a variable next_node to store current.next. Then set current.next to prev. Move prev to current and current to next_node. After the loop, set head to prev.

DSA Python

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

head = Node('Task1')
second = Node('Task2')
third = Node('Task3')
fourth = Node('Task4')

head.next = second
second.next = third
third.next = fourth

prev = None
current = head

# Reverse the linked list iteratively
# Your code here

Hint

Inside the loop, save the next node, reverse the link, then move prev and current forward. After the loop, prev will be the new head.

4

Print the reversed linked list

Create a variable temp and set it to head. Use a while loop that runs while temp is not None. Inside the loop, print temp.data followed by ' -> ' without a newline. Move temp to temp.next. After the loop, print 'None'.

DSA Python

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

head = Node('Task1')
second = Node('Task2')
third = Node('Task3')
fourth = Node('Task4')

head.next = second
second.next = third
third.next = fourth

prev = None
current = head

while current is not None:
    next_node = current.next
    current.next = prev
    prev = current
    current = next_node

head = prev

# Print the reversed linked list
# Your code here

Hint

Use a temporary pointer to go through the reversed list and print each node's data followed by an arrow. End with 'None'.