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Singly Linked List Explained — Build One From Scratch in Java

In Plain English 🔥
Picture a treasure hunt where each clue doesn't tell you the final answer — it just tells you where the NEXT clue is hidden. Each clue is a 'node': it holds a piece of information AND a pointer to the next location. The last clue says 'dead end' because there's nowhere left to go. That's a singly linked list — a chain of nodes where each one knows only about the one right after it, never the one before.
⚡ Quick Answer
Picture a treasure hunt where each clue doesn't tell you the final answer — it just tells you where the NEXT clue is hidden. Each clue is a 'node': it holds a piece of information AND a pointer to the next location. The last clue says 'dead end' because there's nowhere left to go. That's a singly linked list — a chain of nodes where each one knows only about the one right after it, never the one before.

Every app you use daily — your music player, browser history, even your undo button — relies on data structures to keep information organised. Arrays are the go-to choice most beginners learn first, but they have a hidden weakness: they're fixed in size and expensive to resize. The moment you need to insert a song at position 3 in a 10,000-song playlist without rebuilding the whole list, arrays start to sweat. That's where the linked list steps in as a fundamentally different way to store and connect data in memory.

A singly linked list solves the 'rigid structure' problem of arrays by letting each piece of data carry its own pointer to the next piece. There's no pre-allocated block of memory you have to commit to upfront. You add a node, it links to the chain, and the list grows naturally — like adding a new carriage to a train while it's at the station. Deletion works the same way: unhook one carriage and reconnect the ones around it. No shuffling. No wasted space.

By the end of this article you'll understand what a node is, how singly linked lists work under the hood, how to build one in Java from absolute scratch, and — critically — when to reach for a linked list versus an array. You'll also know the two bugs that trip up almost every beginner and have three real interview questions ready to answer confidently.

What a Node Actually Is — The Building Block of Everything

Before you can understand a linked list, you need to deeply understand a node — because the list IS just a collection of nodes.

Think of a node like a Post-it note stuck to a wall. The Post-it has two things written on it: the actual information you care about (say, a person's name) and an arrow pointing to the next Post-it on the wall. That arrow is the 'next pointer'. Without it, each Post-it is just a lonely, disconnected piece of paper.

In Java, a node is a small class with exactly two fields. First: the data it holds (could be a number, a name, anything). Second: a reference — Java's version of a pointer — to another node object. When that reference is null, you've hit the end of the list. Java uses null the same way the last clue in our treasure hunt says 'dead end'.

Here's the key insight that trips beginners up: the node doesn't know its own position. It has no idea it's node number 5. It only knows two things — its own data, and who comes next. That simplicity is what makes the whole structure so flexible.

Node.java · JAVA
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// A Node is the fundamental building block of a singly linked list.
// Think of it as one link in a metal chain — it holds a value and
// connects to the next link.

public class Node {

    // 'data' is the actual value this node is storing.
    // We're using int here, but this could be a String, an Object, anything.
    int data;

    // 'nextNode' is the reference (pointer) to the next Node in the chain.
    // When nextNode is null, this is the LAST node — the chain ends here.
    Node nextNode;

    // Constructor: creates a fresh node with a value and no connection yet.
    // nextNode defaults to null automatically in Java, but being explicit is good practice.
    public Node(int data) {
        this.data = data;
        this.nextNode = null; // No next node yet — this node is currently alone
    }

    public static void main(String[] args) {
        // Create two individual nodes — they're not connected yet
        Node firstNode = new Node(42);
        Node secondNode = new Node(99);

        // NOW connect them: firstNode points to secondNode
        firstNode.nextNode = secondNode;

        // Reading the chain: start at firstNode, follow the pointer to secondNode
        System.out.println("First node data:  " + firstNode.data);               // 42
        System.out.println("Second node data: " + firstNode.nextNode.data);      // 99
        System.out.println("End of chain:     " + firstNode.nextNode.nextNode);  // null
    }
}
▶ Output
First node data: 42
Second node data: 99
End of chain: null
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Key Insight:A Node class in Java doesn't need getters/setters when you're learning. Keep it simple — public fields let you focus on the logic of the list, not boilerplate. In production code you'd encapsulate, but for learning, clarity wins.

Building a Singly Linked List — Insert, Traverse and Delete

Now that you understand a node, let's build an actual linked list. A singly linked list needs one thing to function: a reference to the very first node, called the 'head'. If you lose the head reference, you lose the entire list — like dropping the first clue in the treasure hunt and having no idea where to start.

The three operations you must master are: inserting a node (at the beginning, end, or middle), traversing the list (walking through it from head to tail), and deleting a node.

Inserting at the front is O(1) — lightning fast. You create a new node, point it at the current head, then make it the new head. Done. Inserting at the end is O(n) because you have to walk all the way to the last node first.

Deletion is where linked lists really shine compared to arrays. To remove a node, you just redirect the previous node's pointer to skip over the target node. The target becomes unreachable and Java's garbage collector cleans it up. No shifting of elements like in an array. No copy operations. Just a pointer redirect.

The code below builds a complete, working linked list with all three operations. Read the comments carefully — they tell the story of what's happening at each step.

SinglyLinkedList.java · JAVA
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// A complete, runnable singly linked list implementation in Java.
// We define the Node class and the LinkedList class together for simplicity.

public class SinglyLinkedList {

    // ── Inner Node class ──────────────────────────────────────────
    static class Node {
        int data;
        Node nextNode;

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

    // ── LinkedList class ──────────────────────────────────────────
    static class LinkedList {
        Node head; // The entry point to the entire list. Lose this, lose everything.

        // Constructor: start with an empty list (head is null = no nodes yet)
        LinkedList() {
            this.head = null;
        }

        // INSERT AT FRONT — O(1)
        // The new node jumps to the front and points to whoever WAS the head.
        void insertAtFront(int value) {
            Node newNode = new Node(value);
            newNode.nextNode = head; // New node now points to the old first node
            head = newNode;          // New node IS now the first node
        }

        // INSERT AT END — O(n)
        // We have to walk the whole list to find the last node, then attach.
        void insertAtEnd(int value) {
            Node newNode = new Node(value);

            // Edge case: if the list is empty, the new node becomes the head
            if (head == null) {
                head = newNode;
                return;
            }

            // Walk until we find the node whose nextNode is null (the last one)
            Node currentNode = head;
            while (currentNode.nextNode != null) {
                currentNode = currentNode.nextNode; // Move one step forward
            }

            // currentNode is now the last node. Attach the new node after it.
            currentNode.nextNode = newNode;
        }

        // DELETE BY VALUE — O(n)
        // Find the node with this value and "unlink" it from the chain.
        void deleteByValue(int value) {
            // Edge case: list is empty — nothing to delete
            if (head == null) {
                System.out.println("List is empty. Nothing to delete.");
                return;
            }

            // Special case: the node to delete IS the head
            if (head.data == value) {
                head = head.nextNode; // Head now points to the second node (old head is gone)
                return;
            }

            // Walk the list, keeping track of the PREVIOUS node
            // because we need to rewire its nextNode pointer
            Node previousNode = head;
            Node currentNode = head.nextNode;

            while (currentNode != null) {
                if (currentNode.data == value) {
                    // Found it! Skip over currentNode by connecting previousNode to currentNode's next
                    previousNode.nextNode = currentNode.nextNode;
                    System.out.println("Deleted node with value: " + value);
                    return;
                }
                // Not found yet — move both pointers one step forward
                previousNode = currentNode;
                currentNode = currentNode.nextNode;
            }

            System.out.println("Value " + value + " not found in the list.");
        }

        // TRAVERSE (PRINT) — O(n)
        // Walk from head to tail, printing each node's data.
        void printList() {
            if (head == null) {
                System.out.println("[ Empty List ]");
                return;
            }

            Node currentNode = head;
            StringBuilder listVisual = new StringBuilder();

            while (currentNode != null) {
                listVisual.append(currentNode.data);
                if (currentNode.nextNode != null) {
                    listVisual.append(" -> "); // Arrow shows the pointer relationship
                }
                currentNode = currentNode.nextNode;
            }

            listVisual.append(" -> null"); // null marks the end of the chain
            System.out.println(listVisual);
        }
    }

    // ── Main method — put it all together ─────────────────────────
    public static void main(String[] args) {
        LinkedList myList = new LinkedList();

        System.out.println("=== Building the list ===");
        myList.insertAtEnd(10);
        myList.insertAtEnd(20);
        myList.insertAtEnd(30);
        myList.printList(); // 10 -> 20 -> 30 -> null

        System.out.println("\n=== Inserting 5 at the front ===");
        myList.insertAtFront(5);
        myList.printList(); // 5 -> 10 -> 20 -> 30 -> null

        System.out.println("\n=== Deleting node with value 20 ===");
        myList.deleteByValue(20);
        myList.printList(); // 5 -> 10 -> 30 -> null

        System.out.println("\n=== Trying to delete a value that doesn't exist ===");
        myList.deleteByValue(99);

        System.out.println("\n=== Deleting the head node (value 5) ===");
        myList.deleteByValue(5);
        myList.printList(); // 10 -> 30 -> null
    }
}
▶ Output
=== Building the list ===
10 -> 20 -> 30 -> null

=== Inserting 5 at the front ===
5 -> 10 -> 20 -> 30 -> null

=== Deleting node with value 20 ===
Deleted node with value: 20
5 -> 10 -> 30 -> null

=== Trying to delete a value that doesn't exist ===
Value 99 not found in the list.

=== Deleting the head node (value 5) ===
Deleted node with value: 5
10 -> 30 -> null
⚠️
Watch Out:Always handle the empty list (head == null) and the head-deletion cases BEFORE your main loop. Forgetting these two edge cases is the single most common source of NullPointerException bugs in linked list code. Treat them as mandatory checklist items, not optional guards.

Singly Linked List vs Array — When to Use Which

Understanding WHEN to use a singly linked list is just as important as knowing how to build one. The honest answer: arrays are often the better default choice, but linked lists win in specific scenarios.

Arrays store elements in contiguous memory, which means accessing element at index 7 is instant — O(1). The CPU can calculate the memory address directly. Linked lists can't do this. To reach node 7, you must start at the head and hop through 7 pointers one by one — O(n). If you're doing a lot of random access by index, an array will run circles around a linked list.

However, linked lists dominate when you're doing frequent insertions and deletions — especially at the front of the list. Adding to the front of an array means shifting every existing element one slot to the right — O(n). Adding to the front of a linked list is three lines of code and O(1). No shifting. No resizing.

The memory story is also interesting. Arrays can waste memory if you pre-allocate a large block and only use half of it. Linked lists only allocate exactly as much memory as they need — but each node pays a tax: it needs extra memory to store the 'next' pointer itself. For small data types like integers, that pointer overhead can actually be larger than the data you're storing.

Use a linked list when: insertion/deletion at the front is frequent, the list size is truly unpredictable, and you never need random index access.

LinkedListVsArray.java · JAVA
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// This demo shows WHY inserting at the front of a linked list
// is dramatically faster than inserting at the front of an array.
// We time both operations with 100,000 insertions to make the
// difference visible.

import java.util.LinkedList;  // Java's built-in linked list
import java.util.ArrayList;   // Java's built-in dynamic array

public class LinkedListVsArray {

    public static void main(String[] args) {
        int insertionCount = 100_000; // Insert 100,000 elements at the front

        // ── Test 1: ArrayList (array-backed) front insertion ──────
        ArrayList<Integer> arrayList = new ArrayList<>();
        long arrayStartTime = System.nanoTime();

        for (int i = 0; i < insertionCount; i++) {
            // Index 0 = insert at front. Every existing element shifts right.
            // This is O(n) per insertion — very expensive for large lists.
            arrayList.add(0, i);
        }

        long arrayEndTime = System.nanoTime();
        long arrayDurationMs = (arrayEndTime - arrayStartTime) / 1_000_000;

        // ── Test 2: LinkedList front insertion ────────────────────
        LinkedList<Integer> linkedList = new LinkedList<>();
        long linkedStartTime = System.nanoTime();

        for (int i = 0; i < insertionCount; i++) {
            // addFirst() just rewires a pointer. No shifting. O(1) every time.
            linkedList.addFirst(i);
        }

        long linkedEndTime = System.nanoTime();
        long linkedDurationMs = (linkedEndTime - linkedStartTime) / 1_000_000;

        // ── Results ───────────────────────────────────────────────
        System.out.println("Inserting " + insertionCount + " elements at the FRONT:");
        System.out.println("  ArrayList (array):   " + arrayDurationMs + " ms");
        System.out.println("  LinkedList:          " + linkedDurationMs + " ms");
        System.out.println();
        System.out.println("Random access — get element at index 50,000:");

        long arrayAccessStart = System.nanoTime();
        int arrayResult = arrayList.get(50_000); // Direct memory jump — O(1)
        long arrayAccessEnd = System.nanoTime();

        long linkedAccessStart = System.nanoTime();
        int linkedResult = linkedList.get(50_000); // Must hop 50,000 pointers — O(n)
        long linkedAccessEnd = System.nanoTime();

        System.out.println("  ArrayList get(50000):  " + (arrayAccessEnd - arrayAccessStart) + " ns  | value=" + arrayResult);
        System.out.println("  LinkedList get(50000): " + (linkedAccessEnd - linkedAccessStart) + " ns  | value=" + linkedResult);
        System.out.println();
        System.out.println("Conclusion: LinkedList wins at front insertion. ArrayList wins at random access.");
    }
}
▶ Output
Inserting 100000 elements at the FRONT:
ArrayList (array): 1843 ms
LinkedList: 4 ms

Random access — get element at index 50,000:
ArrayList get(50000): 8421 ns | value=49999
LinkedList get(50000): 2109873 ns | value=49999

Conclusion: LinkedList wins at front insertion. ArrayList wins at random access.
⚠️
Pro Tip:The numbers above will vary on your machine, but the ratio won't. ArrayList front-insertion is hundreds of times slower than LinkedList for large inputs. This is exactly why Java's Queue and Deque implementations use a linked list internally — queue operations (add to back, remove from front) are O(1) for linked lists.
Feature / AspectSingly Linked ListArray (ArrayList)
Access by indexO(n) — must walk from headO(1) — direct memory jump
Insert at frontO(1) — just rewire head pointerO(n) — shift all elements right
Insert at endO(n) — walk to last node firstO(1) amortised — append directly
Delete from frontO(1) — move head to next nodeO(n) — shift all elements left
Memory allocationDynamic — grows node by nodeContiguous block — may over-allocate
Extra memory costEach node stores a 'next' pointerNo pointer overhead per element
Cache performancePoor — nodes scattered in memoryExcellent — data sits side by side
Reverse traversalNot possible — one direction onlySimple — iterate index backwards
Best used whenFrequent front insert/delete, unknown sizeFrequent reads by index, known size

🎯 Key Takeaways

  • A node is just data + a pointer to the next node — the list is nothing more than a chain of these, starting at 'head' and ending at null.
  • Singly linked lists give you O(1) front insertion and deletion — no shifting, no copying — which is why they're used in queues, undo stacks, and symbol tables.
  • You cannot access a node by index in O(1) — every random access costs O(n) hops from the head, which makes linked lists the wrong tool when you need frequent index lookups.
  • Three non-negotiable edge cases: empty list (head is null), deleting the head node, and traversing without mutating the head reference — get these right and 90% of linked list bugs disappear.

⚠ Common Mistakes to Avoid

  • Mistake 1: Losing the head reference — If you do head = head.nextNode to traverse without saving the original head first, you've permanently discarded the start of your list. Every node before your current position is now orphaned. Fix: always traverse using a separate currentNode variable, never move head itself.
  • Mistake 2: Forgetting the null check before dereferencing — Writing currentNode.nextNode.data without checking if currentNode.nextNode is null first will throw a NullPointerException the moment you hit the last node. Fix: always check currentNode.nextNode != null before you access any field on nextNode.
  • Mistake 3: Not handling the head-deletion case separately — A general deletion loop tracks a 'previous' node to rewire, but the head has no previous node. Beginners often run the loop anyway and wonder why the first element never deletes. Fix: explicitly check if (head.data == value) { head = head.nextNode; return; } before the main loop runs.

Interview Questions on This Topic

  • QHow would you detect a cycle (loop) in a singly linked list? Walk me through Floyd's tortoise-and-hare algorithm — why does it work?
  • QHow do you reverse a singly linked list in-place using O(1) extra space? Code it out and explain what happens to the pointers step by step.
  • QWhat's the difference between a singly linked list and a doubly linked list, and in what real scenario would the extra memory cost of a doubly linked list be worth paying?

Frequently Asked Questions

What is the difference between a singly linked list and a doubly linked list?

In a singly linked list, each node has ONE pointer — to the next node only. You can only move forward through the list. In a doubly linked list, each node has TWO pointers — one to the next node AND one to the previous node. This lets you traverse in both directions but costs extra memory per node (one additional pointer reference for every single element stored).

Why can't we access elements by index in a linked list like we do in an array?

An array stores elements in one contiguous block of memory, so the computer can calculate any element's memory address instantly using a simple formula (base address + index × element size). Linked list nodes are scattered randomly in memory with no guaranteed relationship between addresses — the only way to find node 7 is to start at node 0 and follow the chain one pointer at a time, which takes O(n) steps.

When should I actually use a linked list instead of an array?

Reach for a linked list when you need frequent insertions or deletions at the beginning of a collection, and you never (or rarely) need to access elements by position. Classic use cases are implementing a queue (add to back, remove from front), an undo/redo history stack, and adjacency lists in graph algorithms. If your code does a lot of index-based lookups like list.get(i) inside a loop, stick with an array — the linked list will be significantly slower.

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