Creating a Singly Linked List from Scratch in DSA Rust - Code It Step by Step

Creating a Singly Linked List from Scratch

📖 Scenario: Imagine you want to keep a list of your favorite fruits in order. You want to add fruits one by one and see the list grow. A singly linked list is a simple way to do this, where each fruit points to the next one.

🎯 Goal: You will build a singly linked list in Rust from scratch. You will create nodes, link them, and print the list to see the fruits in order.

📋 What You'll Learn

Create a Node struct with a value and a link to the next node

Create a LinkedList struct to manage the list

Add a method to append new nodes to the list

Add a method to print all values in the list

💡 Why This Matters

🌍 Real World

Linked lists are used in many software systems to manage ordered data where insertion and deletion happen frequently.

💼 Career

Understanding linked lists helps in technical interviews and is fundamental for building efficient data structures in system programming.

1

Create the Node struct

Create a Node struct with a field value of type String and a field next of type Option>.

DSA Rust

# Your code here

Hint

Use Option> for the next field to allow linking to the next node or none.

2

Create the LinkedList struct

Create a LinkedList struct with a field head of type Option> to store the start of the list.

DSA Rust

struct Node {
    value: String,
    next: Option>,
}

# Your code here

Hint

The head points to the first node or is None if the list is empty.

3

Add append method to LinkedList

Add an impl LinkedList block with a method append(&mut self, value: String) that adds a new node with the given value at the end of the list.

DSA Rust

struct Node {
    value: String,
    next: Option>,
}

struct LinkedList {
    head: Option>,
}

# Your code here

Hint

Use match on self.head.as_mut() to check if the list is empty or not. Traverse to the end to add the new node.

4

Add print method and display the list

Add a method print_list(&self) inside impl LinkedList that prints all node values separated by -> and ends with None. Then create a main function that creates a LinkedList, appends "Apple", "Banana", and "Cherry", and calls print_list().

DSA Rust

struct Node {
    value: String,
    next: Option>,
}

struct LinkedList {
    head: Option>,
}

impl LinkedList {
    fn append(&mut self, value: String) {
        let new_node = Box::new(Node { value, next: None });
        match self.head.as_mut() {
            None => self.head = Some(new_node),
            Some(mut current) => {
                while let Some(next_node) = current.next.as_mut() {
                    current = next_node;
                }
                current.next = Some(new_node);
            }
        }
    }

    # Your code here
}

fn main() {
    # Your code here
}

Hint

Traverse the list using as_ref() to print each node's value. In main, create the list, append fruits, and print.

Practice

1. What is the main purpose of a Singly Linked List in data structures?

easy

A. To store elements in a random order without any links

B. To store elements in a tree-like hierarchical structure

C. To store elements with direct access by index like an array

D. To store elements in a linear order where each element points to the next

Solution

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

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike arrays, singly linked lists do not provide direct index access and unlike trees, they have a single linear chain of nodes.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option D

4. Quick Check:

   Singly Linked List = Linear chain with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing linked lists with arrays having direct index access
• Thinking linked lists are hierarchical like trees
• Assuming nodes can point backward in singly linked lists

2. Which of the following is the correct way to define a node struct for a singly linked list in Rust?

easy

A. struct Node { value: i32, next: &Node }

B. struct Node { value: i32, next: Node }

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

D. struct Node { value: i32, next: Vec<Node> }

Solution

1. Step 1: Understand Rust ownership and pointers

   Rust requires heap allocation for recursive types like linked lists, so Box is used to own the next node.

2. Step 2: Analyze each option

   struct Node { value: i32, next: Option } uses Option allowing for a nullable next pointer, which is correct. struct Node { value: i32, next: Node } is invalid because it causes infinite size. struct Node { value: i32, next: &Node } uses a reference without lifetime, which is invalid. struct Node { value: i32, next: Vec<Node> } uses a vector which is not suitable for singly linked list next pointer.

3. Final Answer:

   struct Node { value: i32, next: Option } -> Option C

4. Quick Check:

   Use Option for next pointer in Rust [OK]

Hint: Use Option for nullable next node in Rust [OK]

Common Mistakes:

• Using direct Node type causes infinite size error
• Using references without lifetimes for next pointer
• Using Vec<Node> instead of a single next node pointer

3. What will be the printed output after running this Rust code that creates a singly linked list with values 1, 2, 3 and prints them?

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

fn main() {
    let third = Node { value: 3, next: None };
    let second = Node { value: 2, next: Some(Box::new(third)) };
    let first = Node { value: 1, next: Some(Box::new(second)) };

    let mut current = &first;
    while let Some(ref next_node) = current.next {
        println!("{} ->", current.value);
        current = next_node;
    }
    println!("{} -> null", current.value);
}

medium

A. 1 -> 2 -> null

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

C. 3 -> 2 -> 1 -> null

D. Compilation error due to borrowing

Solution

1. Step 1: Trace the linked list creation

   Nodes are created in reverse order: third (3, None), second (2, Some(third)), first (1, Some(second)). So the list is 1 -> 2 -> 3 -> null.

2. Step 2: Follow the print loop

   The loop prints current.value followed by '->' while current.next is Some. It prints 1 ->, then 2 ->, then exits loop at 3 (next is None) and prints 3 -> null.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option B

4. Quick Check:

   Prints nodes in order from first to last [OK]

Hint: Trace node links from first to last for output [OK]

Common Mistakes:

• Printing nodes in reverse order by mistake
• Missing the last node print with 'null'
• Confusing Option with direct node causing errors

4. Identify the error in this Rust code snippet that tries to create a singly linked list node:

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

fn main() {
    let node1 = Node { value: 1, next: None };
    let node2 = Node { value: 2, next: node1 };
}

medium

A. next field must be Option, but node1 is Node type

B. Missing semicolon after node2 declaration

C. Cannot move node1 because it is not boxed inside Option

D. No error, code compiles fine

Solution

1. Step 1: Check the type of next field

   The next field expects an Option, but node1 is a Node, not wrapped in Box or Option.

2. Step 2: Understand Rust type requirements

   Rust requires the next pointer to be boxed and wrapped in Option to handle ownership and nullability. Directly assigning node1 causes a type mismatch.

3. Final Answer:

   next field must be Option, but node1 is Node type -> Option A

4. Quick Check:

   next must be Option not Node [OK]

Hint: Match next field type exactly: Option [OK]

Common Mistakes:

• Assigning Node directly instead of Option
• Ignoring Rust ownership rules for recursive types
• Assuming semicolon missing causes error here

5. You want to implement a method append to add a new node at the end of a singly linked list in Rust. Which approach correctly handles ownership and mutability to traverse and append the node?

hard

A. Use a mutable reference to Option and loop until next is None, then set next to Some(new_node)

B. Use an immutable reference to Node and assign next directly without mutability

C. Clone the entire list and add the new node at the front

D. Use unsafe pointers to bypass Rust ownership rules

Solution

1. Step 1: Understand Rust ownership and mutability for linked lists

   To modify the list, you need a mutable reference to the next pointer (Option) to traverse and update it safely.

2. Step 2: Analyze each option

   Use a mutable reference to Option and loop until next is None, then set next to Some(new_node) correctly uses mutable references to traverse until the end and then sets next to Some(new_node). Use an immutable reference to Node and assign next directly without mutability uses immutable reference which cannot modify. Clone the entire list and add the new node at the front clones the list which is inefficient and not appending at end. Use unsafe pointers to bypass Rust ownership rules uses unsafe code which is discouraged.

3. Final Answer:

   Use a mutable reference to Option and loop until next is None, then set next to Some(new_node) -> Option A

4. Quick Check:

   Mutable reference to next pointer needed for append [OK]

Hint: Traverse with mutable Option reference to append [OK]

Common Mistakes:

• Using immutable references preventing modification
• Trying to assign next without Option or Box
• Using unsafe code unnecessarily