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

Reverse a Singly Linked List Iterative

📖 Scenario: You are working with a simple list of tasks represented as a singly linked list. Sometimes, you want to see the tasks in reverse order to plan your day backward.

🎯 Goal: Build a program that creates a singly linked list of 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 exactly these tasks in order: "Wake up", "Brush teeth", "Eat breakfast", "Go to work"

Add a variable to help reverse the list iteratively

Write the iterative logic to reverse the singly linked list

Print the reversed list in the format: Task1 -> Task2 -> Task3 -> Task4 -> null

💡 Why This Matters

🌍 Real World

Reversing linked lists is useful in many applications like undo operations, navigation history, and task scheduling where order needs to be reversed.

💼 Career

Understanding linked list reversal is a common interview question and fundamental for software engineers working with data structures.

1

Create the Singly Linked List

Define a struct called Node with fields task string and next *Node. Then create four nodes with tasks "Wake up", "Brush teeth", "Eat breakfast", and "Go to work" linked in that order. Finally, create a variable head pointing to the first node.

DSA Go

// Define Node struct and create linked list nodes
// Your code here

Hint

Start by defining the Node struct. Then create nodes from last to first so you can link them easily.

2

Add Variables for Reversal

Add two variables var prev *Node and current *Node and set current = head. These will help you reverse the list iteratively.

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func main() {
	node4 := &Node{task: "Go to work", next: nil}
	node3 := &Node{task: "Eat breakfast", next: node4}
	node2 := &Node{task: "Brush teeth", next: node3}
	node1 := &Node{task: "Wake up", next: node2}

	head := node1

	// Add variables prev and current for reversal
	// Your code here
}

Hint

Initialize prev as nil and current as head to start reversal.

3

Reverse the Linked List Iteratively

Write a for loop that runs while current != nil. Inside the loop, save current.next in nextNode, then set current.next = prev. Move prev to current and current to nextNode. After the loop, set head = prev.

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func main() {
	node4 := &Node{task: "Go to work", next: nil}
	node3 := &Node{task: "Eat breakfast", next: node4}
	node2 := &Node{task: "Brush teeth", next: node3}
	node1 := &Node{task: "Wake up", next: node2}

	head := node1

	var prev *Node
	current := head

	// Write the iterative reversal loop here
	// Your code here
}

Hint

Remember to save the next node before changing links, then move pointers forward.

4

Print the Reversed Linked List

Write a for loop to traverse the reversed list starting from head. Print each task followed by " -> ". After the loop, print "null" to show the end of the list.

DSA Go

package main

import "fmt"

type Node struct {
	task string
	next *Node
}

func main() {
	node4 := &Node{task: "Go to work", next: nil}
	node3 := &Node{task: "Eat breakfast", next: node4}
	node2 := &Node{task: "Brush teeth", next: node3}
	node1 := &Node{task: "Wake up", next: node2}

	head := node1

	var prev *Node
	current := head

	for current != nil {
		nextNode := current.next
		current.next = prev
		prev = current
		current = nextNode
	}

	head = prev

	// Print the reversed list
	// Your code here
}

Hint

Use a loop to print each task followed by " -> ". After the loop, print "null" to mark the end.

Practice

1. What is the main purpose of the iterative method to reverse a singly linked list?

easy

A. To sort the linked list in ascending order

B. To create a new linked list with reversed elements

C. To delete all nodes and recreate them in reverse order

D. To change the direction of the links between nodes one by one

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes pointing to the next node only, so reversing means changing these pointers.

2. Step 2: Purpose of iterative reversal

   The iterative method changes the next pointer of each node to point to the previous node, reversing the list in place.

3. Final Answer:

   To change the direction of the links between nodes one by one -> Option D

4. Quick Check:

   Reversing means changing links direction [OK]

Hint: Reversing means flipping links, not creating new nodes [OK]

Common Mistakes:

• Thinking it creates a new list instead of reversing in place
• Confusing reversal with sorting
• Assuming nodes are deleted and recreated

2. Which of the following Go code snippets correctly initializes pointers for iterative reversal of a singly linked list?

easy

A. var prev = head; current := nil

B. var prev *Node = nil; current := head

C. prev := head; current := head.next

D. var prev *Node = head; current := head

Solution

1. Step 1: Initialization for reversal

   We start with prev as nil because the new tail will point to nil, and current at head to traverse the list.

2. Step 2: Check each option

   var prev *Node = nil; current := head correctly sets prev to nil and current to head. Others incorrectly assign prev or current.

3. Final Answer:

   var prev *Node = nil; current := head -> Option B

4. Quick Check:

   Prev starts nil, current at head [OK]

Hint: Prev starts nil, current starts at head [OK]

Common Mistakes:

• Setting prev to head initially
• Setting current to nil at start
• Mixing pointer assignments

3. What will be the output linked list after running this Go code to reverse a singly linked list?

type Node struct {
    Val  int
    Next *Node
}

func reverseList(head *Node) *Node {
    var prev *Node = nil
    current := head
    for current != nil {
        nextTemp := current.Next
        current.Next = prev
        prev = current
        current = nextTemp
    }
    return prev
}

// Input list: 1 -> 2 -> 3 -> nil
// Call: reverseList(head)

medium

A. 3 -> 2 -> 1 -> nil

B. 1 -> 2 -> 3 -> nil

C. nil -> 3 -> 2 -> 1

D. 3 -> 1 -> 2 -> nil

Solution

1. Step 1: Trace the reversal loop

   Initially prev=nil, current=1. Each iteration reverses current.Next to prev and moves pointers forward.

2. Step 2: Final reversed list

   After loop, prev points to node with Val=3, linked as 3->2->1->nil.

3. Final Answer:

   3 -> 2 -> 1 -> nil -> Option A

4. Quick Check:

   Reversed list order is 3,2,1 [OK]

Hint: Follow pointer changes step-by-step [OK]

Common Mistakes:

• Confusing original and reversed order
• Misunderstanding pointer updates
• Assuming list remains unchanged

4. Identify the error in this Go code snippet for reversing a singly linked list iteratively:

func reverseList(head *Node) *Node {
    var prev *Node = head
    current := head
    for current != nil {
        nextTemp := current.Next
        current.Next = prev
        prev = current
        current = nextTemp
    }
    return prev
}

medium

A. prev should be initialized to nil, not head

B. current should be initialized to nil

C. nextTemp should be initialized outside the loop

D. The function should return current, not prev

Solution

1. Step 1: Check initialization of prev

   Prev must start as nil because the new tail's next is nil. Setting prev=head causes a cycle.

2. Step 2: Effect of wrong prev initialization

   Setting prev=head makes current.Next point to itself on first iteration, causing infinite loop or wrong reversal.

3. Final Answer:

   prev should be initialized to nil, not head -> Option A

4. Quick Check:

   Prev starts nil to avoid cycles [OK]

Hint: Prev must start nil to break old links [OK]

Common Mistakes:

• Initializing prev to head causing cycles
• Returning current instead of prev
• Misplacing nextTemp initialization

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> nil, write the final linked list after applying iterative reversal twice in a row.

hard

A. 40 -> 30 -> 20 -> 10 -> nil

B. nil

C. 10 -> 20 -> 30 -> 40 -> nil

D. 40 -> 10 -> 20 -> 30 -> nil

Solution

1. Step 1: First reversal

   Reversing 10->20->30->40->nil gives 40->30->20->10->nil.

2. Step 2: Second reversal

   Reversing again restores original order: 10->20->30->40->nil.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> nil -> Option C

4. Quick Check:

   Double reversal returns original list [OK]

Hint: Reversing twice returns original list [OK]

Common Mistakes:

• Assuming second reversal changes order differently
• Confusing node values after reversal
• Thinking list becomes empty