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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the Node class's next pointer.

DSA Java

class Node {
    int data;
    Node [1];

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

Drag options to blanks, or click blank then click option'

Alink

Bprev

Cpointer

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using incorrect variable names like 'pointer' or 'link'.

The 'next' pointer in a singly linked list node points to the next node.

2fill in blank

medium

Complete the code to initialize the head of the linked list.

DSA Java

class LinkedList {
    Node head;

    LinkedList() {
        this.[1] = null;
    }
}

Drag options to blanks, or click blank then click option'

Ahead

Btail

Cstart

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing 'tail' instead of 'head'.

Using other names like 'start' or 'root' which are not standard here.

The head pointer points to the first node in the linked list and is initialized to null when the list is empty.

3fill in blank

hard

Fix the error in the method to add a new node at the front of the list.

DSA Java

void addFront(int data) {
    Node newNode = new Node(data);
    newNode.next = [1];
    head = newNode;
}

Drag options to blanks, or click blank then click option'

Ahead

Btail

CnewNode

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Setting newNode.next to null loses the rest of the list.

Using 'tail' or 'newNode' instead of 'head' causes errors.

To add a node at the front, newNode.next should point to the current head.

4fill in blank

hard

Fill both blanks to complete the method that prints all node data in the list.

DSA Java

void printList() {
    Node current = head;
    while (current [1] null) {
        System.out.print(current.data + " -> ");
        current = current.[2];
    }
    System.out.println("null");
}

Drag options to blanks, or click blank then click option'

A!=

B==

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' in the while condition.

Using 'prev' instead of 'next' to move forward.

The loop continues while current is not null, and current moves to current.next each iteration.

5fill in blank

hard

Fill all three blanks to complete the method that counts nodes in the list.

DSA Java

int countNodes() {
    int count = 0;
    Node current = [1];
    while (current [2] null) {
        count[3];
        current = current.next;
    }
    return count;
}

Drag options to blanks, or click blank then click option'

Ahead

B!=

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from null instead of head.

Using '--' to decrement count instead of incrementing.

Using '==' instead of '!=' in the loop condition.

Start from head, loop while current is not null, increment count each time.

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