Why Reverse a Doubly Linked List in DSA Rust?

The Big Idea

What if you could flip a list backward instantly without rewriting it?

The Scenario

Imagine you have a list of your favorite songs saved in order. Now, you want to play them in reverse order. If you try to do this by writing down each song again from the end to the start manually, it will take a lot of time and you might make mistakes.

The Problem

Manually reversing a list means you have to remember each item and rewrite it backwards. This is slow and easy to mess up, especially if the list is long. You might lose some songs or mix up the order.

The Solution

Reversing a doubly linked list lets the computer flip the order by just changing the links between songs. It swaps the directions so the last song becomes the first, without copying or rewriting the whole list. This is fast and safe.

Before vs After

✗ Before

let mut reversed = Vec::new();
for i in (0..list.len()).rev() {
    reversed.push(list[i]);
}

✓ After

fn reverse_list(head: &mut Option<Box<Node>>) {
    let mut current = head.take();
    while let Some(mut node) = current {
        std::mem::swap(&mut node.next, &mut node.prev);
        current = node.prev.take();
        *head = Some(node);
    }
}

What It Enables

This lets you quickly flip the order of items in a list, making it easy to access data backward without extra memory or copying.

Real Life Example

Music apps use this to let you play your playlist backward instantly, without creating a new list or waiting.

Key Takeaways

Manual reversing is slow and error-prone.

Reversing a doubly linked list swaps links efficiently.

This operation saves time and memory for reversing order.

Practice

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

easy

A. To change the order of nodes so the last becomes first and vice versa

B. To delete all nodes from the list

C. To add new nodes at the end of the list

D. To sort the nodes in ascending order

Solution

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

   A doubly linked list has nodes connected both forward and backward.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To change the order of nodes so the last becomes first and vice versa -> Option A

4. Quick Check:

   Reversing means flipping order = C [OK]

Hint: Reversing flips the list order front to back [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Mixing up adding nodes with reversing

2. Which Rust code snippet correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. let temp = node.prev; node.prev = node.next; node.next = temp.next;

B. node.next = node.next; node.prev = node.prev;

C. node.prev = node.next; node.next = node.prev;

D. let temp = node.next; node.next = node.prev; node.prev = temp;

Solution

1. Step 1: Identify correct swapping logic

   Swapping means storing one pointer temporarily, then exchanging values.

2. Step 2: Check each option

   let temp = node.next; node.next = node.prev; node.prev = temp; correctly uses a temp variable to swap next and prev. Others either assign incorrectly or use wrong variables.

3. Final Answer:

   let temp = node.next; node.next = node.prev; node.prev = temp; -> Option D

4. Quick Check:

   Swap pointers with temp variable = D [OK]

Hint: Use a temp variable to swap pointers safely [OK]

Common Mistakes:

• Assigning pointers without temp causes data loss
• Mixing up next and prev in assignments
• Using wrong pointer in swap

3. Given the Rust code below that reverses a doubly linked list, what will be the printed output?

struct Node {
    data: i32,
    prev: Option,
    next: Option,
}

// Assume list is 1 <-> 2 <-> 3
// After reversal, print from head

// Output format: data values separated by spaces

// Output?

medium

A. 3 2 1

B. 1 2 3

C. 2 3 1

D. 3 1 2

Solution

1. Step 1: Understand initial list order

   The list is 1 <-> 2 <-> 3, so printing from head gives 1 2 3.

2. Step 2: Effect of reversal

   Reversing swaps next and prev pointers, so the new head is 3, and order is 3 2 1.

3. Final Answer:

   3 2 1 -> Option A

4. Quick Check:

   Reversed list prints backwards = 3 2 1 [OK]

Hint: Reversed doubly linked list prints nodes backward [OK]

Common Mistakes:

• Printing original order after reversal
• Confusing node data order
• Assuming reversal sorts nodes

4. Identify the bug in this Rust function that attempts to reverse a doubly linked list:

fn reverse(mut head: Option) -> Option {
    let mut current = head;
    while let Some(mut node) = current {
        let temp = node.next;
        node.next = node.prev;
        node.prev = temp;
        current = node.next;
    }
    head
}

medium

A. The loop condition should be while current.is_some()

B. Swapping next and prev pointers is done incorrectly

C. The function does not update head to the new first node after reversal

D. The function should return current instead of head

Solution

1. Step 1: Analyze pointer swapping

   Swapping next and prev is done correctly inside the loop.

2. Step 2: Check head update

   The function returns original head without updating it to the new first node after reversal, so head remains old start.

3. Final Answer:

   The function does not update head to the new first node after reversal -> Option C

4. Quick Check:

   Head must point to new first node after reverse = A [OK]

Hint: Update head to last processed node after reversal [OK]

Common Mistakes:

• Not updating head pointer after reversal
• Returning old head instead of new head
• Incorrect loop condition causing infinite loop

5. You have a doubly linked list with nodes containing integers. After reversing it, you want to find the sum of all nodes that were originally at even positions (2nd, 4th, etc.). Which approach correctly solves this in Rust?

hard

A. Reverse the list, then traverse and sum nodes at odd positions in the reversed list

B. Traverse original list, sum nodes at even positions, then reverse the list

C. Reverse the list, then sum all nodes without position checks

D. Traverse reversed list and sum nodes at even positions

Solution

1. Step 1: Understand position change after reversal

   Reversing changes node positions, so original even positions become different in reversed list.

2. Step 2: Correct approach

   Sum nodes at even positions before reversal to get original even-position sum, then reverse list if needed.

3. Final Answer:

   Traverse original list, sum nodes at even positions, then reverse the list -> Option B

4. Quick Check:

   Sum original even positions before reversing = A [OK]

Hint: Sum before reversing to keep original positions intact [OK]

Common Mistakes:

• Summing positions after reversal which changes order
• Ignoring position changes due to reversal
• Summing all nodes without filtering positions