Reverse a Singly Linked List Iterative in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind reversing a singly linked list iteratively?

We change the direction of each node's pointer one by one, so the list points backward instead of forward.

beginner

In the iterative reversal of a singly linked list, what role does the 'previous' pointer play?

It keeps track of the node that should come before the current node once the link is reversed.

beginner

Why do we need a 'next' pointer during the iterative reversal process?

To remember the next node before changing the current node's pointer, so we don't lose access to the rest of the list.

beginner

What is the time complexity of reversing a singly linked list iteratively?

O(n), where n is the number of nodes, because we visit each node exactly once.

intermediate

Show the Rust code snippet to reverse a singly linked list iteratively.

```rust
fn reverse_list(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    let mut prev = None;
    while let Some(mut current) = head {
        head = current.next.take();
        current.next = prev;
        prev = Some(current);
    }
    prev
}
```

What does the 'prev' pointer represent during iterative reversal of a singly linked list?

A The tail of the original list

B The next node to process

C The head of the original list

D The node after the current node after reversal

The 'prev' pointer always points to the node that will come before the current node after reversal.

Why do we use 'current.next.take()' in Rust when reversing a linked list?

A To move ownership of the next node out safely

B To clone the next node

C To delete the next node

D To borrow the next node immutably

In Rust, 'take()' moves the ownership out, allowing us to safely change pointers without borrowing conflicts.

What happens to the original head pointer after the iterative reversal completes?

A It becomes None

B It points to the new head (original tail)

C It points to the middle node

D It remains unchanged

During reversal, the original head pointer moves through the list and ends as None when the list is fully reversed.

What is the space complexity of reversing a singly linked list iteratively?

A O(log n)

B O(n)

C O(1)

D O(n^2)

The iterative reversal uses only a few pointers, so it uses constant extra space.

Which of these is NOT needed to reverse a singly linked list iteratively?

A A pointer to the previous node

B A stack to store nodes

C A pointer to the next node

D A pointer to the current node

A stack is not needed; the reversal is done by changing pointers directly.

Explain step-by-step how to reverse a singly linked list using iteration.

Think about how you change the direction of each link one by one.

Write the Rust function signature and describe the key steps inside the function to reverse a singly linked list iteratively.

Focus on ownership and pointer changes in Rust.

Practice

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

easy

A. To change the direction of the links so the last node becomes the first

B. To sort the linked list in ascending order

C. To remove duplicate nodes from the list

D. To find the middle node of the list

Solution

1. Step 1: Understand the linked list structure

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

2. Step 2: Identify the effect of reversal

   Reversing changes the direction so the last node points to null and becomes the new head.

3. Final Answer:

   To change the direction of the links so the last node becomes the first -> Option A

4. Quick Check:

   Reversing = changing link directions [OK]

Hint: Reversing flips links, last node becomes first [OK]

Common Mistakes:

• Confusing reversal with sorting
• Thinking reversal removes nodes
• Assuming reversal finds middle node

2. Which of the following Rust code snippets correctly initializes a mutable pointer for reversing a singly linked list iteratively?

easy

A. let mut prev = Some(Node::new());

B. let prev = Some(Node::new());

C. let mut prev = None;

D. let prev = None;

Solution

1. Step 1: Understand pointer initialization for reversal

   We start with a previous pointer set to None because initially no nodes are reversed.

2. Step 2: Check mutability and value

   We need mutable prev to update it during iteration, so 'let mut prev = None;' is correct.

3. Final Answer:

   let mut prev = None; -> Option C

4. Quick Check:

   Mutable prev starts as None [OK]

Hint: Start prev as mutable None to track reversed list [OK]

Common Mistakes:

• Forgetting mut keyword
• Initializing prev with a new node incorrectly
• Using immutable prev when mutation is needed

3. Given the following Rust code snippet for reversing a singly linked list iteratively, what will be the printed output after reversal?

struct Node {
    val: i32,
    next: Option,
}

fn print_list(head: &Option) {
    let mut current = head;
    while let Some(node) = current {
        print!("{} -> ", node.val);
        current = &node.next;
    }
    println!("null");
}

fn reverse_list(mut head: Option) -> Option {
    let mut prev = None;
    while let Some(mut current) = head {
        head = current.next.take();
        current.next = prev;
        prev = Some(current);
    }
    prev
}

fn main() {
    let node3 = Node { val: 3, next: None };
    let node2 = Node { val: 2, next: Some(Box::new(node3)) };
    let node1 = Node { val: 1, next: Some(Box::new(node2)) };
    let head = Some(Box::new(node1));

    let reversed_head = reverse_list(head);
    print_list(&reversed_head);
}

medium

A. null

B. 1 -> 2 -> 3 -> null

C. Compilation error

D. 3 -> 2 -> 1 -> null

Solution

1. Step 1: Trace the reversal process

   The list starts as 1 -> 2 -> 3 -> null. The reversal changes links so the last node (3) becomes head.

2. Step 2: Confirm print output after reversal

   After reversal, print_list prints nodes from new head: 3 -> 2 -> 1 -> null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option D

4. Quick Check:

   Reversed list prints in reverse order [OK]

Hint: Reversed list prints nodes in reverse order [OK]

Common Mistakes:

• Assuming list prints original order after reversal
• Confusing Option C null with empty list
• Thinking code causes compilation error

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

fn reverse_list(mut head: Option) -> Option {
    let mut prev = None;
    while let Some(current) = head {
        head = current.next;
        current.next = prev;
        prev = Some(current);
    }
    prev
}

medium

A. prev should be immutable

B. Missing .take() when assigning head = current.next

C. Loop condition should be while head.is_some()

D. Function should return head instead of prev

Solution

1. Step 1: Understand ownership in Rust linked list

   current.next is an Option, but to move it out, .take() is needed to avoid borrowing errors.

2. Step 2: Identify the error in head assignment

   head = current.next; tries to move without .take(), causing ownership error. Correct is head = current.next.take();

3. Final Answer:

   Missing .take() when assigning head = current.next -> Option B

4. Quick Check:

   Use .take() to move ownership safely [OK]

Hint: Use .take() to move Option ownership in Rust [OK]

Common Mistakes:

• Ignoring Rust ownership rules
• Making prev immutable by mistake
• Changing loop condition unnecessarily

5. You have a singly linked list with nodes containing integer values. You want to reverse the list iteratively and then remove all nodes with even values from the reversed list. Which approach correctly combines these two operations in Rust?

hard

A. First reverse the list iteratively, then iterate over reversed list to filter out even nodes by rebuilding a new list

B. Filter out even nodes first, then reverse the filtered list recursively

C. Reverse the list recursively and remove even nodes during reversal by skipping them

D. Use a stack to push all nodes, pop them to reverse, and skip even nodes while popping

Solution

1. Step 1: Reverse the list iteratively

   Use the iterative reversal method to get the reversed list with all nodes.

2. Step 2: Filter out even nodes from reversed list

   Iterate over reversed list and build a new list skipping nodes with even values.

3. Final Answer:

   First reverse the list iteratively, then iterate over reversed list to filter out even nodes by rebuilding a new list -> Option A

4. Quick Check:

   Reverse then filter by rebuilding list [OK]

Hint: Reverse first, then filter by rebuilding list [OK]

Common Mistakes:

• Trying to filter during reversal causing complexity
• Using recursion when iterative is required
• Using stack unnecessarily increasing space