Reverse a Singly Linked List Iterative in DSA C - 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 in a singly linked list. Sometimes, you want to see the tasks in reverse order. To do this, you will reverse the linked list using an iterative method.

🎯 Goal: Build a C program that creates a singly linked list with exact tasks, then reverses the list iteratively, and finally prints the reversed list.

📋 What You'll Learn

Create a singly linked list with nodes containing integer values 1, 2, 3, 4, 5 in that order

Add a pointer variable to help reverse the list

Implement an iterative function to reverse the singly linked list

Print the reversed linked list values separated by ' -> ' and ending with ' -> NULL'

💡 Why This Matters

🌍 Real World

Reversing linked lists is useful in many applications like undo features, navigation history, and data processing pipelines.

💼 Career

Understanding linked list reversal is a common interview question and helps build strong problem-solving skills in data structures.

1

Create the singly linked list

Define a struct called Node with an int data and a Node* called next. Then create five nodes with data values 1, 2, 3, 4, 5 linked in order. Finally, create a pointer called head that points to the first node.

DSA C

// Define struct Node and create linked list nodes with values 1 to 5
// Link nodes and set head pointer
// Your code here

Hint

Remember to link each node's next pointer to the following node, and set the last node's next to NULL.

2

Add pointers for reversing the list

Inside main, declare three pointers: prev, current, and next. Initialize prev to NULL and current to head. These will help reverse the list iteratively.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));
    Node* node4 = (Node*)malloc(sizeof(Node));
    Node* node5 = (Node*)malloc(sizeof(Node));

    node1->data = 1;
    node2->data = 2;
    node3->data = 3;
    node4->data = 4;
    node5->data = 5;

    node1->next = node2;
    node2->next = node3;
    node3->next = node4;
    node4->next = node5;
    node5->next = NULL;

    Node* head = node1;

    // Declare pointers prev, current, next
    // Initialize prev to NULL and current to head
    // Your code here

    return 0;
}

Hint

These pointers will help you keep track of nodes while reversing the links.

3

Iteratively reverse the linked list

Write a while loop that runs while current is not NULL. Inside the loop, do these steps in order: set next to current->next, set current->next to prev, move prev to current, and move current to next. After the loop, set head to prev.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));
    Node* node4 = (Node*)malloc(sizeof(Node));
    Node* node5 = (Node*)malloc(sizeof(Node));

    node1->data = 1;
    node2->data = 2;
    node3->data = 3;
    node4->data = 4;
    node5->data = 5;

    node1->next = node2;
    node2->next = node3;
    node3->next = node4;
    node4->next = node5;
    node5->next = NULL;

    Node* head = node1;

    Node* prev = NULL;
    Node* current = head;
    Node* next = NULL;

    // Use a while loop to reverse the list iteratively
    // Your code here

    return 0;
}

Hint

Change the direction of each node's next pointer to point to the previous node.

4

Print the reversed linked list

Use a Node* pointer called temp to traverse the reversed list starting from head. Inside a while loop, print the data of each node followed by " -> ". After the loop, print "NULL" and a newline.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));
    Node* node4 = (Node*)malloc(sizeof(Node));
    Node* node5 = (Node*)malloc(sizeof(Node));

    node1->data = 1;
    node2->data = 2;
    node3->data = 3;
    node4->data = 4;
    node5->data = 5;

    node1->next = node2;
    node2->next = node3;
    node3->next = node4;
    node4->next = node5;
    node5->next = NULL;

    Node* head = node1;

    Node* prev = NULL;
    Node* current = head;
    Node* next = NULL;

    while (current != NULL) {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    head = prev;

    // Print the reversed list
    // Your code here

    return 0;
}

Hint

Traverse the list from head and print each node's data followed by ' -> '. End with 'NULL'.

Practice

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

easy

A. To delete all nodes from the linked list

B. To create a new linked list with reversed elements

C. To sort the linked list in ascending 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 connected in one direction only, from head to tail.

2. Step 2: Purpose of iterative reversal

   The iterative method changes the direction of each node's pointer to point to the previous node, reversing the list.

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 reversal creates a new list instead of changing links
• Confusing reversal with sorting
• Assuming nodes are deleted during reversal

2. Which of the following is the correct way to initialize pointers for iterative reversal of a singly linked list in C?

easy

A. struct Node *prev = NULL, *current = head, *next = NULL;

B. struct Node *prev = head, *current = NULL, *next = NULL;

C. struct Node *prev = NULL, *current = NULL, *next = head;

D. struct Node *prev = head, *current = head, *next = head;

Solution

1. Step 1: Identify starting pointers

   For reversal, prev starts as NULL because the new tail points to NULL, current starts at head to traverse.

2. Step 2: Correct initialization

   Next pointer is initialized as NULL and updated inside the loop; so initial next = NULL is correct.

3. Final Answer:

   struct Node *prev = NULL, *current = head, *next = NULL; -> Option A

4. Quick Check:

   Prev NULL, current head to start [OK]

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

Common Mistakes:

• Setting prev to head initially
• Setting current to NULL at start
• Initializing next pointer incorrectly

3. What will be the output linked list after running the iterative reverse function on the list: 1 -> 2 -> 3 -> NULL?

medium

A. 2 -> 3 -> 1 -> NULL

B. 3 -> 2 -> 1 -> NULL

C. NULL -> 3 -> 2 -> 1

D. 1 -> 2 -> 3 -> NULL

Solution

1. Step 1: Trace reversal process

   Initially: 1->2->3->NULL. After reversal, links point backward: 3->2->1->NULL.

2. Step 2: Confirm final head

   The last node (3) becomes the new head, so output is 3->2->1->NULL.

3. Final Answer:

   3 -> 2 -> 1 -> NULL -> Option B

4. Quick Check:

   Reverse order of nodes [OK]

Hint: Reverse means last becomes first [OK]

Common Mistakes:

• Assuming list stays the same after reversal
• Confusing NULL placement
• Mixing node order incorrectly

4. Identify the error in this iterative reversal code snippet for a singly linked list in C:

struct Node* reverse(struct Node* head) {
  struct Node* prev = head;
  struct Node* current = head;
  struct Node* next = NULL;
  while (current != NULL) {
    next = current->next;
    current->next = prev;
    prev = current;
    current = next;
  }
  return prev;
}

medium

A. prev should be initialized to NULL, not head

B. current should be initialized to NULL, not head

C. next pointer is not updated inside the loop

D. Function should return current, not prev

Solution

1. Step 1: Check pointer initialization

   Prev must start as NULL because the new tail node points to NULL, not head.

2. Step 2: Analyze loop pointer updates

   Current and next are correctly updated; only prev initialization is wrong.

3. Final Answer:

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

4. Quick Check:

   Prev NULL to mark new tail [OK]

Hint: Prev must start NULL to avoid cycle [OK]

Common Mistakes:

• Starting prev at head causing cycle
• Not updating next inside loop
• Returning wrong pointer at end

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

hard

A. NULL

B. 40 -> 30 -> 20 -> 10 -> NULL

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

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

Solution

1. Step 1: Reverse the list once

   Original: 10->20->30->40->NULL. After first reversal: 40->30->20->10->NULL.

2. Step 2: Reverse the reversed list again

   Reversing 40->30->20->10->NULL returns it to original: 10->20->30->40->NULL.

3. Final Answer:

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

4. Quick Check:

   Double reversal restores original list [OK]

Hint: Reversing twice returns original list [OK]

Common Mistakes:

• Assuming double reversal changes list order
• Confusing node values after reversal
• Forgetting NULL at list end