Get Length of Linked List in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a linked list?

A linked list is a chain of nodes where each node holds data and a link to the next node. It is like a treasure hunt where each clue leads to the next one.

beginner

How do you find the length of a linked list?

Start at the first node and move through each node one by one, counting as you go, until you reach the end (where there is no next node). The count is the length.

intermediate

Why can't you get the length of a linked list directly like an array?

Because linked lists don't store their size. You must walk through all nodes to count them, unlike arrays which know their size upfront.

intermediate

Rust code snippet: How to traverse a linked list node safely?

Use a loop with an Option reference to the current node. Move to the next node by updating the reference until it becomes None.

beginner

What is the time complexity of getting the length of a linked list?

It is O(n), where n is the number of nodes, because you must visit each node once to count them.

What does each node in a linked list contain?

A Only a link to the next node

B Only data

C Data and a link to the next node

D Data and the total length

Each node holds data and a pointer to the next node, forming the chain.

How do you know when you have reached the end of a linked list?

A When the data is zero

B When the next node is None

C When the node has no data

D When the length counter reaches zero

The last node points to None, indicating the end of the list.

What is the first step to get the length of a linked list?

A Start counting from the head node

B Check the length property

C Jump to the last node

D Reverse the list

You start at the head and count nodes one by one.

What Rust type is commonly used to represent the next node in a linked list?

A i32

B Vec<Node>

C String

D Option<Box<Node>>

Option> allows for nullable ownership of the next node.

What is the time complexity to get the length of a linked list with n nodes?

A O(n)

B O(1)

C O(log n)

D O(n^2)

You must visit each node once, so it takes linear time.

Explain step-by-step how to find the length of a linked list.

Think of walking through a chain counting each link.

Describe why linked lists require traversal to find their length, unlike arrays.

Compare how arrays and linked lists store size information.

Practice

1. What does the function to get the length of a linked list typically do?

easy

A. Counts the number of nodes by traversing from head to end

B. Returns the value of the first node

C. Deletes all nodes in the list

D. Finds the maximum value in the list

Solution

1. Step 1: Understand the linked list traversal

   The function starts at the head node and moves through each node one by one.

2. Step 2: Count each node visited

   For every node visited, the function increments a counter until it reaches the end (null).

3. Final Answer:

   Counts the number of nodes by traversing from head to end -> Option A

4. Quick Check:

   Length = number of nodes visited [OK]

Hint: Traverse and count nodes until null [OK]

Common Mistakes:

• Returning the value of the first node instead of count
• Not handling empty list (null head)
• Trying to count without traversal

2. Which Rust code snippet correctly initializes a linked list node with value 5 and next pointer None?

easy

A. let node = Node::new(5);

B. let node = Node(5, None);

C. let node = Node { value: 5, next: None };

D. let node = Node { value: 5 };

Solution

1. Step 1: Check struct initialization syntax

   Rust structs are initialized with field names and values inside braces.

2. Step 2: Match field names and order

   let node = Node { value: 5, next: None }; uses correct field names and syntax. Tuple syntax requires tuple struct definition. Node::new(5) assumes an impl with new method. Omitting next field causes compile error.

3. Final Answer:

   let node = Node { value: 5, next: None }; -> Option C

4. Quick Check:

   Struct init uses field: value pairs [OK]

Hint: Use struct field names with braces for initialization [OK]

Common Mistakes:

• Using tuple syntax for struct without tuple struct definition
• Missing field names in initialization
• Omitting required fields like next

3. What is the output of this Rust code that calculates the length of a linked list?

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

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while let Some(node) = current {
        count += 1;
        current = &node.next;
    }
    count
}

fn main() {
    let node3 = Node { value: 3, next: None };
    let node2 = Node { value: 2, next: Some(Box::new(node3)) };
    let node1 = Node { value: 1, next: Some(Box::new(node2)) };
    let head = Some(Box::new(node1));
    println!("{}", length(&head));
}

medium

A. 1

B. 2

C. 0

D. 3

Solution

1. Step 1: Understand the linked list structure

   There are three nodes linked: node1 -> node2 -> node3 -> None.

2. Step 2: Trace the length function

   The function counts nodes by traversing from head. It visits node1, node2, and node3, incrementing count each time.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   3 nodes counted [OK]

Hint: Count nodes by following next pointers until None [OK]

Common Mistakes:

• Counting only head node
• Not dereferencing Option correctly
• Confusing Option with Box

4. Identify the error in this Rust function to get linked list length:

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while current.is_some() {
        count += 1;
        current = current.unwrap().next;
    }
    count
}

medium

A. Using unwrap() on &Option causes a move error

B. Loop condition should be current.is_none()

C. Count should start at 1, not 0

D. Function should return i32, not usize

Solution

1. Step 1: Analyze unwrap usage on reference

   Calling unwrap() on &Option tries to move out of a borrowed reference, which is not allowed.

2. Step 2: Correct way to access next node

   Use pattern matching or while let Some(node) = current to borrow without moving.

3. Final Answer:

   Using unwrap() on &Option causes a move error -> Option A

4. Quick Check:

   unwrap() moves ownership, not allowed on references [OK]

Hint: Use pattern matching, avoid unwrap on references [OK]

Common Mistakes:

• Using unwrap() on borrowed Option causing move error
• Wrong loop condition logic
• Incorrect return type for length

5. Given a linked list where each node contains an integer, how would you modify the length function to count only nodes with even values?

fn length_even(head: &Option) -> usize {
    // Your code here
}

hard

A. Use recursion without checking node values

B. Traverse list, increment count only if node.value % 2 == 0

C. Return length() / 2 assuming half are even

D. Count all nodes, then subtract odd values count

Solution

1. Step 1: Traverse the linked list nodes

   Start from head and move through each node until None.

2. Step 2: Check each node's value for evenness

   Increment count only if node.value % 2 == 0 (even number).

3. Step 3: Return the count of even-valued nodes

   This gives the length of nodes with even values.

4. Final Answer:

   Traverse list, increment count only if node.value % 2 == 0 -> Option B

5. Quick Check:

   Count only even nodes by checking value % 2 == 0 [OK]

Hint: Count nodes while checking even condition [OK]

Common Mistakes:

• Counting all nodes without filtering
• Assuming half nodes are even without check
• Using recursion without base case