Singly linked list structure in Data Structures Theory - Mini Project: Build & Apply

Singly linked list structure

📖 Scenario: You want to understand how a singly linked list works by building its basic structure step-by-step.Imagine a chain of boxes where each box holds a value and a pointer to the next box.

🎯 Goal: Build a simple singly linked list structure with nodes that hold values and links to the next node.

📋 What You'll Learn

Create a node structure with a value and a next pointer

Create a variable to hold the head of the list

Link nodes together to form the list

Add a final pointer to mark the end of the list

💡 Why This Matters

🌍 Real World

Singly linked lists are used in many software systems to manage ordered data efficiently, such as in music playlists or undo history.

💼 Career

Understanding linked lists is fundamental for software developers and engineers working with data structures and algorithms.

1

Create the node structure

Create a class called Node with an __init__ method that takes a parameter value and sets self.value = value and self.next = None.

Data Structures Theory

# Create the Node class with __init__ method
# Your code here

Hint

Think of each node as a box holding a value and a link to the next box.

2

Create the head of the list

Create a variable called head and set it to None to represent an empty list.

Data Structures Theory

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

# Create the head variable and set it to None
# Your code here

Hint

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

3

Create and link nodes

Create two nodes called node1 with value 10 and node2 with value 20. Set head to node1 and link node1.next to node2.

Data Structures Theory

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

head = None

# Create node1 and node2, link them, and set head
# Your code here

Hint

Link the first node to the second by setting node1.next to node2.

4

Mark the end of the list

Set node2.next to None to mark the end of the list.

Data Structures Theory

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None

head = None

node1 = Node(10)
node2 = Node(20)
head = node1
node1.next = node2

# Set node2.next to None to end the list
# Your code here

Hint

The last node's next pointer should always be None to show the list ends here.

Practice

1. What does each node in a singly linked list contain?

easy

A. Only data without any pointers

B. Data and pointers to both next and previous nodes

C. Data and a pointer to the next node

D. Only a pointer to the previous node

Solution

1. Step 1: Understand node structure in singly linked list

   Each node stores some data and a pointer to the next node only, not to the previous one.

2. Step 2: Compare options with definition

   Data and a pointer to the next node matches the definition exactly. Data and pointers to both next and previous nodes describes a doubly linked list. Options C and D miss the pointer to the next node.

3. Final Answer:

   Data and a pointer to the next node -> Option C

4. Quick Check:

   Node = data + next pointer [OK]

Hint: Remember: singly linked means one pointer per node [OK]

Common Mistakes:

• Confusing singly linked with doubly linked list
• Thinking nodes have no pointers
• Assuming pointer to previous node exists

2. Which of the following is the correct way to represent the head of a singly linked list in code?

easy

A. head = None

B. head = head.next

C. head = head.prev

D. head = data

Solution

1. Step 1: Understand the head pointer initialization

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

2. Step 2: Analyze options

   head = None correctly initializes head as None for an empty list. head = head.next tries to move head to next node, which is invalid if list is empty. head = head.prev uses prev pointer which does not exist in singly linked list. head = data assigns data to head, which is incorrect.

3. Final Answer:

   head = None -> Option A

4. Quick Check:

   Empty list head = None [OK]

Hint: Head starts as None when list is empty [OK]

Common Mistakes:

• Using prev pointer in singly linked list
• Assigning data directly to head
• Moving head before list has nodes

3. Given the following singly linked list nodes:

node1 = Node(data=10, next=node2)
node2 = Node(data=20, next=node3)
node3 = Node(data=30, next=None)
head = node1

What is the data value of head.next.next?

medium

A. 20

B. 30

C. 10

D. None

Solution

1. Step 1: Trace the linked nodes from head

   head points to node1 with data 10. head.next points to node2 with data 20. head.next.next points to node3 with data 30.

2. Step 2: Identify the data at head.next.next

   The data at node3 is 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   head.next.next.data = 30 [OK]

Hint: Follow next pointers step-by-step [OK]

Common Mistakes:

• Confusing node positions
• Assuming next of last node is not None
• Mixing data values with node references

4. Identify the error in this singly linked list insertion code snippet:

new_node = Node(data=5)
new_node.next = head
head = new_node.next

medium

A. new_node.next should be None

B. No error, code is correct

C. Data assignment to new_node is missing

D. head should be assigned to new_node, not new_node.next

Solution

1. Step 1: Understand insertion at head

   To insert new_node at the start, new_node.next should point to current head, then head should be updated to new_node.

2. Step 2: Analyze the code error

   The code sets new_node.next = head correctly, but then sets head = new_node.next, which points to old head, not new_node. This loses the new node as head.

3. Final Answer:

   head should be assigned to new_node, not new_node.next -> Option D

4. Quick Check:

   head = new_node after insertion [OK]

Hint: After linking, head must point to new_node [OK]

Common Mistakes:

• Assigning head to new_node.next instead of new_node
• Forgetting to link new_node.next to old head
• Not updating head pointer after insertion

5. You want to reverse a singly linked list with nodes:

head -> node1 -> node2 -> node3 -> None

Which sequence of pointer changes correctly reverses the list?

hard

A. Change each node's next pointer to the previous node, then update head to last node

B. Swap data values between nodes without changing pointers

C. Set head to None and create a new list with reversed data

D. Change each node's next pointer to None

Solution

1. Step 1: Understand reversal of singly linked list

   Reversing means changing each node's next pointer to point to the previous node instead of the next.

2. Step 2: Identify correct pointer update sequence

   Change each node's next pointer to the previous node, then update head to last node correctly describes reversing pointers and updating head to the last node. Swap data values between nodes without changing pointers only swaps data, not pointers. Set head to None and create a new list with reversed data discards the list unnecessarily. Change each node's next pointer to None breaks links without reversing.

3. Final Answer:

   Change each node's next pointer to the previous node, then update head to last node -> Option A

4. Quick Check:

   Reverse = reverse pointers + update head [OK]

Hint: Reverse pointers, then reset head to last node [OK]

Common Mistakes:

• Swapping data instead of pointers
• Not updating head after reversal
• Breaking links without reconnecting