Creating a Singly Linked List from Scratch in DSA Java - 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 from scratch in Java. You will create nodes, link them, and print the list to see the fruits in order.

📋 What You'll Learn

Create a Node class with a string data and a next pointer

Create a LinkedList class with a head pointer

Add nodes to the linked list

Print the linked list in order

💡 Why This Matters

🌍 Real World

Singly linked lists are used in many applications like music playlists, undo features in apps, and managing tasks in order.

💼 Career

Understanding linked lists is fundamental for software developers, especially for roles involving data structures, algorithms, and system design.

1

Create the Node class

Create a class called Node with two members: a String data and a Node next. Add a constructor that takes a String data and sets next to null.

DSA Java

// Your code here to create the Node class

Hint

Think of each node as a box holding a fruit name and a pointer to the next box.

2

Create the LinkedList class with head

Create a class called LinkedList with a member Node head initialized to null. This will be the start of your list.

DSA Java

class Node {
    String data;
    Node next;

    Node(String data) {
        this.data = data;
        this.next = null;
    }
}

// Your code here to create the LinkedList class with head

Hint

The head is like the first fruit in your basket. If it's empty, the basket is empty.

3

Add a method to insert nodes at the end

Inside the LinkedList class, create a method called add(String data) that creates a new Node with the given data and adds it to the end of the list. If the list is empty, set head to the new node. Otherwise, traverse to the last node and link the new node.

DSA Java

class Node {
    String data;
    Node next;

    Node(String data) {
        this.data = data;
        this.next = null;
    }
}

class LinkedList {
    Node head;

    LinkedList() {
        this.head = null;
    }

    // Add the add method below
    // Your code here
}

Hint

Think of walking through the list until you find the last fruit, then add the new fruit after it.

4

Print the linked list

Inside the LinkedList class, create a method called printList() that prints all the node data in order separated by -> and ends with null. Then, in the main method, create a LinkedList, add these fruits in order: "Apple", "Banana", "Cherry", and call printList().

DSA Java

class Node {
    String data;
    Node next;

    Node(String data) {
        this.data = data;
        this.next = null;
    }
}

class LinkedList {
    Node head;

    LinkedList() {
        this.head = null;
    }

    void add(String data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
        } else {
            Node current = head;
            while (current.next != null) {
                current = current.next;
            }
            current.next = newNode;
        }
    }

    // Add the printList method below
    // Your code here
}

public class Main {
    public static void main(String[] args) {
        LinkedList list = new LinkedList();
        // Add "Apple", "Banana", "Cherry" to the list
        // Call printList to display the list
        // Your code here
    }
}

Hint

Print each fruit followed by an arrow until you reach the end, then print null.

Practice

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

easy

A. To store elements in a sequence where each element points to the next

B. To store elements with keys and values for fast lookup

C. To store elements in a sorted array for quick access

D. To store elements in a tree structure with parent and child nodes

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 or trees, singly linked lists do not store keys or have parent-child relationships; they only link sequentially.

3. Final Answer:

   To store elements in a sequence where each element points to the next -> Option A

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes link one after another [OK]

Common Mistakes:

• Confusing linked list with arrays or hash maps
• Thinking linked list stores key-value pairs
• Assuming linked list nodes have multiple pointers

2. Which of the following is the correct way to declare a node class for a singly linked list in Java?

easy

A. class Node { int data; Node left; Node right; }

B. class Node { int data; Node next; }

C. class Node { int data; Node prev; }

D. class Node { int data; int next; }

Solution

1. Step 1: Identify the node structure for singly linked list

   A singly linked list node contains data and a reference to the next node only.

2. Step 2: Check each option

   class Node { int data; Node next; } has data and next pointer; class Node { int data; Node prev; } has prev (used in doubly linked list); class Node { int data; Node left; Node right; } has left and right (used in trees); class Node { int data; int next; } incorrectly uses int for next pointer.

3. Final Answer:

   class Node { int data; Node next; } -> Option B

4. Quick Check:

   Node = data + next pointer [OK]

Hint: Node must have 'next' pointer, not 'prev' or others [OK]

Common Mistakes:

• Using 'prev' pointer instead of 'next'
• Using primitive type for next pointer
• Confusing tree node with linked list node

3. What will be the output of the following code snippet that creates a singly linked list with nodes 10, 20, 30 and prints the list?

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

public class LinkedList {
  Node head;
  public void add(int data) {
    Node newNode = new Node(data);
    if (head == null) {
      head = newNode;
    } else {
      Node temp = head;
      while (temp.next != null) {
        temp = temp.next;
      }
      temp.next = newNode;
    }
  }
  public void printList() {
    Node temp = head;
    while (temp != null) {
      System.out.print(temp.data + " -> ");
      temp = temp.next;
    }
    System.out.print("null");
  }

  public static void main(String[] args) {
    LinkedList list = new LinkedList();
    list.add(10);
    list.add(20);
    list.add(30);
    list.printList();
  }
}

medium

A. 30 -> 20 -> 10 -> null

B. null

C. 10 -> 20 -> null

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Understand the add method behavior

   The add method appends new nodes at the end of the list, preserving insertion order.

2. Step 2: Trace the printList method

   It prints nodes from head to last, showing 10, then 20, then 30, followed by null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option D

4. Quick Check:

   Adding nodes appends at end, print shows 10->20->30->null [OK]

Hint: Add appends at end; print shows order added [OK]

Common Mistakes:

• Assuming add inserts at head instead of tail
• Printing nodes in reverse order
• Missing the final 'null' in output

4. Identify the error in the following code that attempts to add a node to the singly linked list:

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

public class LinkedList {
  Node head;
  public void add(int data) {
    Node newNode = new Node(data);
    if (head == null) {
      head = newNode;
    } else {
      Node temp = head;
      while (temp != null) {
        temp = temp.next;
      }
      temp.next = newNode;
    }
  }
}

medium

A. The while loop condition should be 'temp.next != null' instead of 'temp != null'

B. The newNode should be created outside the add method

C. The head should be initialized to newNode inside the else block

D. The Node class should have a 'prev' pointer

Solution

1. Step 1: Analyze the while loop condition

   The loop currently runs while temp != null, which moves temp to null and loses reference to last node.

2. Step 2: Correct the loop to stop at last node

   The loop should run while temp.next != null to stop at the last node, so we can set temp.next = newNode.

3. Final Answer:

   The while loop condition should be 'temp.next != null' instead of 'temp != null' -> Option A

4. Quick Check:

   Loop stops at last node using temp.next != null [OK]

Hint: Loop to last node uses 'temp.next != null' [OK]

Common Mistakes:

• Looping until temp is null loses last node reference
• Creating newNode outside add method is unnecessary
• Adding 'prev' pointer in singly linked list node

5. You want to create a singly linked list from scratch that supports adding nodes at the head and printing the list. Which code snippet correctly implements adding at the head?

hard

A. public void addAtHead(int data) { Node newNode = new Node(data); head = newNode; newNode.next = head; }

B. public void addAtHead(int data) { Node newNode = new Node(data); head.next = newNode; head = newNode; }

C. public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; }

D. public void addAtHead(int data) { Node newNode = new Node(data); newNode = head; head = newNode; }

Solution

1. Step 1: Understand adding at head logic

   To add at head, new node's next should point to current head, then head updated to new node.

2. Step 2: Check each option's correctness

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } correctly sets newNode.next = head and then head = newNode. Others incorrectly assign or reverse pointers.

3. Final Answer:

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } -> Option C

4. Quick Check:

   Add at head = newNode.next = head; head = newNode [OK]

Hint: New head points to old head, then update head [OK]

Common Mistakes:

• Assigning head.next before updating head
• Reversing pointer assignments
• Overwriting newNode reference incorrectly