Creating a Singly Linked List from Scratch in DSA Java - Step-by-Step

Mental Model

A singly linked list is a chain of nodes where each node points to the next one, forming a line.

Analogy: Imagine a treasure hunt where each clue points you to the next clue, one after another, until the treasure is found.

head -> [node1] -> [node2] -> [node3] -> null

Dry Run Walkthrough

Input: Create a singly linked list with values 1, 2, 3 in that order

Goal: Build a linked list that holds 1 -> 2 -> 3 -> null

Step 1: Create first node with value 1 and set it as head

head -> [1] -> null

Why: We need a starting point for the list

Step 2: Create second node with value 2 and link it after first node

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

Why: Add next element by linking from first node

Step 3: Create third node with value 3 and link it after second node

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

Why: Continue linking nodes to build the list

Result:

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

Annotated Code

DSA Java

public class SinglyLinkedList {
    static class Node {
        int data;
        Node next;
        Node(int data) {
            this.data = data;
            this.next = null;
        }
    }

    Node head;

    public void append(int data) {
        Node newNode = new Node(data); // create new node
        if (head == null) {
            head = newNode; // first node becomes head
            return;
        }
        Node current = head;
        while (current.next != null) {
            current = current.next; // move to last node
        }
        current.next = newNode; // link new node at end
    }

    public void printList() {
        Node current = head;
        while (current != null) {
            System.out.print(current.data + " -> ");
            current = current.next; // move forward
        }
        System.out.println("null");
    }

    public static void main(String[] args) {
        SinglyLinkedList list = new SinglyLinkedList();
        list.append(1);
        list.append(2);
        list.append(3);
        list.printList();
    }
}

Node newNode = new Node(data); // create new node

create a new node with given data

if (head == null) { head = newNode; // first node becomes head return; }

if list empty, new node is head

while (current.next != null) { current = current.next; // move to last node }

traverse to last node

current.next = newNode; // link new node at end

link new node after last node

while (current != null) { System.out.print(current.data + " -> "); current = current.next; // move forward }

print all nodes in order

Output

1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(n) because appending requires traversing all n nodes to find the end

Space: O(n) because each new node uses extra space proportional to number of elements

vs Alternative: Compared to arrays, linked lists allow easy insertion without shifting elements but require traversal to append

Edge Cases

empty list

head is null, first append sets head to new node

DSA Java

if (head == null) {
    head = newNode; // first node becomes head
    return;
}

single element list

append links new node after head

DSA Java

while (current.next != null) {
    current = current.next; // move to last node
}

When to Use This Pattern

When you need a simple chain of elements with easy insertion at the end, use a singly linked list to avoid shifting elements.

Common Mistakes

Mistake: Not setting head when list is empty causes null pointer errors
Fix: Check if head is null and assign new node as head before traversing

Mistake: Forgetting to set newNode.next to null can cause cycles
Fix: Always initialize newNode.next to null in node constructor

Summary

Creates a chain of nodes where each points to the next, forming a singly linked list.

Use when you want a simple list structure that grows dynamically without shifting elements.

The key is linking each new node at the end by traversing from the head.

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