Why Reverse a Doubly Linked List in DSA C++?

The Big Idea

What if you could flip a chain of items instantly without breaking a single link?

The Scenario

Imagine you have a paper chain made of links, and you want to flip it so the last link becomes the first. Doing this by hand means unhooking each link and reattaching it in reverse order.

The Problem

Manually reversing the chain is slow and easy to mess up. You might lose track of links or break the chain. It takes a lot of time and careful handling.

The Solution

Reversing a doubly linked list swaps the direction of links automatically by changing pointers. This way, the last element becomes the first without breaking the chain or losing any links.

Before vs After

✗ Before

Node* reverseList(Node* head) {
  // Manually detach and reattach each node
  // Complex and error-prone
}

✓ After

Node* reverseList(Node* head) {
  Node* current = head;
  Node* temp = nullptr;
  while (current != nullptr) {
    temp = current->prev;
    current->prev = current->next;
    current->next = temp;
    current = current->prev;
  }
  if (temp != nullptr) head = temp->prev;
  return head;
}

What It Enables

This lets you quickly flip the order of data in a list, enabling efficient backward and forward traversals.

Real Life Example

Think of a music playlist where you want to play songs in reverse order without creating a new list. Reversing the doubly linked list of songs makes this easy.

Key Takeaways

Manual reversal is slow and risky.

Pointer swapping reverses the list safely and quickly.

Useful for reversing order without extra memory.

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To make the last node the first and the first node the last

B. To delete all nodes from the list

C. To sort the list in ascending order

D. To convert it into a singly linked list

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes with pointers to both previous and next nodes.

2. Step 2: Define reversing operation

   Reversing means swapping the direction so the last node becomes the first and vice versa.

3. Final Answer:

   To make the last node the first and the first node the last -> Option A

4. Quick Check:

   Reverse = swap start and end [OK]

Hint: Reversing swaps head and tail pointers [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Assuming it changes list type

2. Which of the following is the correct way to swap the next and prev pointers of a node curr in C++?

easy

A. curr->next = curr->prev; curr->prev = curr->next;

B. std::swap(curr->next, curr->prev);

C. curr->prev = curr->next; curr->next = curr->prev;

D. curr->next = curr->next; curr->prev = curr->prev;

Solution

1. Step 1: Understand pointer swapping

   Swapping two pointers requires a temporary storage or a swap function to avoid overwriting.

2. Step 2: Use std::swap for safe swapping

   std::swap exchanges values safely without losing data.

3. Final Answer:

   std::swap(curr->next, curr->prev); -> Option B

4. Quick Check:

   Use std::swap to exchange pointers [OK]

Hint: Use std::swap to exchange two pointers safely [OK]

Common Mistakes:

• Assigning one pointer to another without temp causes data loss
• Swapping pointers manually without temp variable
• Confusing assignment with swapping

3. What will be the output of the following code snippet after reversing the doubly linked list?

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

// List: 1 <-> 2 <-> 3
Node* head = new Node{1, nullptr, nullptr};
Node* second = new Node{2, nullptr, head};
head->next = second;
Node* third = new Node{3, nullptr, second};
second->next = third;

// Reverse function
Node* reverse(Node* head) {
  Node* curr = head;
  Node* temp = nullptr;
  while (curr != nullptr) {
    temp = curr->prev;
    curr->prev = curr->next;
    curr->next = temp;
    curr = curr->prev;
  }
  if (temp != nullptr) head = temp->prev;
  return head;
}

head = reverse(head);

// Print list
Node* tempPrint = head;
while (tempPrint != nullptr) {
  std::cout << tempPrint->data << " ";
  tempPrint = tempPrint->next;
}

medium

A. 2 3 1

B. 1 2 3

C. 3 2 1

D. 3 1 2

Solution

1. Step 1: Trace pointer swaps in reverse function

   Each node swaps its next and prev pointers, then moves to original next (now prev).

2. Step 2: Update head to new first node

   After loop, head is updated to last processed node's prev, which is the original tail.

3. Final Answer:

   3 2 1 -> Option C

4. Quick Check:

   Reversed list prints from tail to head [OK]

Hint: After reversing, head points to original tail [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev pointers
• Printing from old head instead of new

4. Identify the error in the following reverse function for a doubly linked list and select the fix.

Node* reverse(Node* head) {
  Node* curr = head;
  Node* temp = nullptr;
  while (curr != nullptr) {
    temp = curr->prev;
    curr->prev = curr->next;
    curr->next = temp;
    curr = curr->next;  // Error here
  }
  if (temp != nullptr) head = temp->prev;
  return head;
}

medium

A. Change 'head = temp->prev;' to 'head = temp->next;'

B. Remove the line 'temp = curr->prev;'

C. Add 'curr = curr->prev;' before the loop

D. Change 'curr = curr->next;' to 'curr = curr->prev;'

Solution

1. Step 1: Analyze pointer movement in loop

   After swapping next and prev, moving curr to curr->next is wrong because next and prev are swapped.

2. Step 2: Correct pointer update

   We must move curr to curr->prev (original next) after swapping.

3. Final Answer:

   Change 'curr = curr->next;' to 'curr = curr->prev;' -> Option D

4. Quick Check:

   Move curr along swapped pointers [OK]

Hint: After swapping, move curr to prev pointer [OK]

Common Mistakes:

• Moving curr to next instead of prev after swap
• Not updating head correctly
• Ignoring pointer swap effects

5. Given a doubly linked list with nodes containing integers, how would you modify the reverse function to also count how many nodes were reversed and return that count along with the new head?

hard

A. Modify the function to return a pair, increment count inside the loop, and return both.

B. Add a global variable to count nodes and return only the head.

C. Count nodes before reversal and return the count without changing the function.

D. Use recursion to reverse and count nodes simultaneously without changing return type.

Solution

1. Step 1: Change function return type

   Use std::pair to return both new head and count.

2. Step 2: Count nodes during reversal

   Initialize count=0; increment count inside the reversal loop for each node processed.

3. Step 3: Return both values

   Return std::make_pair(new_head, count) after reversal completes.

4. Final Answer:

   Modify the function to return a pair, increment count inside the loop, and return both. -> Option A

5. Quick Check:

   Return both head and count using pair [OK]

Hint: Return pair with head and count, increment count in loop [OK]

Common Mistakes:

• Using global variables instead of return values
• Counting nodes before reversal only
• Trying recursion without changing return type