Pop Using Linked List Node in DSA Java

Mental Model

Pop removes the first item from a linked list by moving the start pointer to the next node.

Analogy: Imagine a line of people holding hands. To remove the first person, you just let go of their hand and hold the next person's hand instead.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, pop once

Goal: Remove the first node and update the head to the next node

Step 1: head moves from node 1 to node 2

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

Why: We remove the first node by making head point to the second node

Step 2: node 1 is disconnected and ready for garbage collection

head -> 2 -> 3 -> null

Why: The old first node is no longer referenced and effectively removed

Result:

head -> 2 -> 3 -> null

Annotated Code

DSA Java

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

    Node head;

    // Pop removes the first node and returns its value
    public int pop() {
        if (head == null) {
            throw new RuntimeException("List is empty");
        }
        int poppedValue = head.data; // save data to return
        head = head.next; // move head to next node, removing first
        return poppedValue;
    }

    // Helper to add node at end for setup
    public void append(int data) {
        if (head == null) {
            head = new Node(data);
            return;
        }
        Node curr = head;
        while (curr.next != null) {
            curr = curr.next;
        }
        curr.next = new Node(data);
    }

    // Print list in format: 1 -> 2 -> 3 -> null
    public void printList() {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.data + " -> ");
            curr = curr.next;
        }
        System.out.println("null");
    }

    public static void main(String[] args) {
        LinkedList list = new LinkedList();
        list.append(1);
        list.append(2);
        list.append(3);
        list.printList();
        int popped = list.pop();
        System.out.println("Popped: " + popped);
        list.printList();
    }
}

if (head == null) { throw new RuntimeException("List is empty"); }

check if list is empty to avoid errors

int poppedValue = head.data;

save data from first node to return later

head = head.next;

move head pointer to next node, removing first node

return poppedValue;

return the removed node's data

Output

1 -> 2 -> 3 -> null

Popped: 1

2 -> 3 -> null

Complexity Analysis

Time: O(1) because we only change the head pointer without traversing the list

Space: O(1) because no extra space is used besides a few variables

vs Alternative: Compared to removing from the end which requires O(n) traversal, popping from the front is faster and simpler

Edge Cases

empty list

throws RuntimeException to indicate no nodes to pop

DSA Java

if (head == null) { throw new RuntimeException("List is empty"); }

single node list

head moves to null, list becomes empty after pop

DSA Java

head = head.next;

When to Use This Pattern

When you need to remove the first element quickly from a linked list, use the pop pattern by moving the head pointer to the next node.

Common Mistakes

Mistake: Trying to remove the first node by traversing the list or modifying next pointers incorrectly
Fix: Simply move the head pointer to head.next to remove the first node

Mistake: Not checking if the list is empty before popping
Fix: Add a check for head == null and handle it properly

Summary

Removes the first node from a linked list by moving the head pointer.

Use when you want to quickly remove the front element of a linked list.

The key is that popping just changes the head pointer to the next node.

Practice

1. What does the pop() operation do in a linked list when implemented using nodes?

easy

A. Removes and returns the first node of the linked list

B. Adds a new node at the end of the linked list

C. Removes the last node without returning it

D. Returns the size of the linked list

Solution

1. Step 1: Understand pop operation in linked list context

   The pop operation removes the top element, which is the first node in a linked list acting like a stack.

2. Step 2: Identify what is returned

   Pop returns the value of the removed first node and updates the head pointer to the next node.

3. Final Answer:

   Removes and returns the first node of the linked list -> Option A

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the head node and returns it [OK]

Common Mistakes:

• Thinking pop removes the last node
• Confusing pop with push operation
• Assuming pop returns list size

2. Which of the following is the correct Java syntax to pop a node from the head of a singly linked list?

easy

A. head = head.next; return head.data;

B. Node temp = head; head = head.next; return temp.data;

C. Node temp = head.next; head = temp; return head.data;

D. return head.next.data; head = head.next;

Solution

1. Step 1: Check correct order of operations

   We must save the current head node in a temp variable before moving head to next.

2. Step 2: Return the data of the removed node

   Return the data from the saved temp node, not the new head.

3. Final Answer:

   Node temp = head; head = head.next; return temp.data; -> Option B

4. Quick Check:

   Save head, move head, return old data [OK]

Hint: Save head before moving it to next [OK]

Common Mistakes:

• Returning data from new head instead of old
• Not saving old head before moving
• Returning data before updating head

3. Given the linked list nodes: 10 -> 20 -> 30 -> null, what is the list after one pop operation using this code?

Node pop() {
  if (head == null) return null;
  Node temp = head;
  head = head.next;
  return temp;
}

medium

A. 30 -> null

B. 10 -> 20 -> 30 -> null

C. 20 -> 30 -> null

D. null

Solution

1. Step 1: Identify the node removed by pop()

   Pop removes the head node which has value 10.

2. Step 2: Update head to next node

   Head moves from 10 to 20, so list becomes 20 -> 30 -> null.

3. Final Answer:

   20 -> 30 -> null -> Option C

4. Quick Check:

   Pop removes first node 10 [OK]

Hint: Pop removes first node, head moves next [OK]

Common Mistakes:

• Assuming pop removes last node
• Not updating head pointer
• Returning old list unchanged

4. What is the error in this pop method for a singly linked list?

Node pop() {
  if (head == null) return null;
  Node temp = head.next;
  head = temp;
  return temp;
}

medium

A. It skips the first node and removes the second node instead

B. It causes a null pointer exception when list is empty

C. It returns the old head node instead of new head

D. It does not update the head pointer

Solution

1. Step 1: Analyze which node is assigned to temp

   Temp is assigned head.next, so the first node is skipped.

2. Step 2: Understand effect on head and return

   Head moves to second node, so first node is never removed; second node is removed instead.

3. Final Answer:

   It skips the first node and removes the second node instead -> Option A

4. Quick Check:

   Pop must remove first node, not second [OK]

Hint: Pop must save head, not head.next [OK]

Common Mistakes:

• Assigning temp to head.next instead of head
• Returning the new head instead of the old head
• Returning wrong node

5. You have a linked list: 5 -> 15 -> 25 -> 35 -> null. You perform two pop operations using this method:

Node pop() {
  if (head == null) return null;
  Node temp = head;
  head = head.next;
  return temp;
}

What is the state of the list after these two pops?

hard

A. null

B. 15 -> 25 -> 35 -> null

C. 35 -> null

D. 25 -> 35 -> null

Solution

1. Step 1: First pop removes 5

   After first pop, head moves from 5 to 15, list is 15 -> 25 -> 35 -> null.

2. Step 2: Second pop removes 15

   After second pop, head moves from 15 to 25, list is 25 -> 35 -> null.

3. Final Answer:

   25 -> 35 -> null -> Option D

4. Quick Check:

   Two pops remove first two nodes [OK]

Hint: Each pop removes head, moves head next [OK]

Common Mistakes:

• Removing last nodes instead of first
• Not updating head after each pop
• Confusing pop with push