Get Length of Linked List in DSA Rust

Mental Model

Count how many nodes are in the linked list by moving from the start to the end, one node at a time.

Analogy: Like counting people in a line by starting at the front and moving down the line until no one is left.

head -> [1] -> [2] -> [3] -> null
          ↑

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null

Goal: Find how many nodes are in the list

Step 1: start at head node with value 1

head -> [1] -> [2] -> [3] -> null
          ↑ count=0

Why: We begin counting from the first node

Step 2: count node 1 and move to next node

head -> [1] -> [2] -> [3] -> null
               ↑ count=1

Why: Count increases by one for the current node

Step 3: count node 2 and move to next node

head -> [1] -> [2] -> [3] -> null
                    ↑ count=2

Why: Keep counting nodes until we reach the end

Step 4: count node 3 and move to next node (null)

head -> [1] -> [2] -> [3] -> null
                         ↑ count=3

Why: Last node counted before reaching null

Step 5: stop counting as next node is null

head -> [1] -> [2] -> [3] -> null
                              count=3

Why: No more nodes to count

Result:

head -> [1] -> [2] -> [3] -> null
count = 3

Annotated Code

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    val: i32,
    next: Option<Rc<RefCell<Node>>>,
}

impl Node {
    fn new(val: i32) -> Rc<RefCell<Node>> {
        Rc::new(RefCell::new(Node { val, next: None }))
    }
}

fn get_length(head: Option<Rc<RefCell<Node>>>) -> usize {
    let mut count = 0;
    let mut current = head;
    while let Some(node) = current {
        count += 1; // count current node
        current = node.borrow().next.clone(); // move to next node
    }
    count
}

fn main() {
    // Create linked list: 1 -> 2 -> 3 -> null
    let node1 = Node::new(1);
    let node2 = Node::new(2);
    let node3 = Node::new(3);
    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());

    let length = get_length(Some(node1));
    println!("Length of linked list: {}", length);
}

while let Some(node) = current {

loop over nodes until no more nodes

count += 1; // count current node

increment count for each visited node

current = node.borrow().next.clone(); // move to next node

advance current pointer to next node

Output

Length of linked list: 3

Complexity Analysis

Time: O(n) because we visit each node once to count

Space: O(1) because we only use a few variables for counting

vs Alternative: Alternative is to store length in the list structure for O(1) access but requires extra updates on insert/delete

Edge Cases

empty list (head is None)

returns 0 because there are no nodes

DSA Rust

while let Some(node) = current {

single node list

returns 1 as only one node is counted

DSA Rust

count += 1; // count current node

When to Use This Pattern

When asked to find how many elements are in a linked list, use a simple traversal counting nodes until the end.

Common Mistakes

Mistake: Forgetting to move to the next node inside the loop causing infinite loop
Fix: Add 'current = node.borrow().next.clone();' inside the loop to advance pointer

Mistake: Starting count at 1 instead of 0 causing off-by-one error
Fix: Initialize count to 0 and increment inside the loop for each node

Summary

Counts the number of nodes in a linked list by traversing from head to end.

Use when you need to know the size of a linked list without stored length.

The key is to move node by node and count until you reach the end (null).

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