Linked Lists

A chain of boxes, each one knows only where to find the next

🗺️In the Real World...

Imagine a treasure hunt. You find the first clue, it doesn't tell you where ALL the other clues are. It only tells you where the NEXT clue is. You find that one, it tells you the next, and so on. That's a linked list. A chain of boxes where each box knows exactly ONE thing: where to find the next box.

🔗

Stage 1 / 7

What is a Linked List?

Each box = a value + a pointer to the next box

HEAD↓

→

NULL

👆 Click any node to inspect its contents.

Every node holds exactly 2 things:

val, the data (any value)

next, a pointer to the next node (or None at the end)

Like a treasure hunt, each clue only tells you where to find the next clue.

The two things every node holds:

val, the data you care about (any value)

next, a pointer to the next node (or None at the end)

There is no "jump to node 5" shortcut. There is no length counter built in. Just: start at head, follow the chain.

Now translate what you just saw into Pythonpython

class Node:
    def __init__(self, val):
        self.val = val   # ← the value
        self.next = None # ← the pointer (empty to start)

# Build: 1 → 2 → 3 → None
a = Node(1)
b = Node(2)
c = Node(3)
a.next = b   # 1 points to 2
b.next = c   # 2 points to 3

# Traverse it, the same pattern you stepped through in the visual
curr = a
while curr:
    print(curr.val)  # → 1, then 2, then 3
    curr = curr.next

Prepend is O(1) because you update exactly 2 pointers, always.
new_node.next = head ← point new node at current headhead = new_node ← make new node the head
Those 2 lines work whether the list has 4 nodes or 4 million. The rest of the list is completely untouched. That's the power of pointers.

🤔Quick Check

A linked list has 1,000 nodes. You prepend a new node. How many existing nodes did you visit?

🤔Quick Check

Why can't you do linked_list[5] to get the 6th element?

Practice Exercises

0/3 solved

Understand First

Step 1 / 3

Walk the Chain

You just stepped through a traversal in the interactive visual. Now do it in code, same exact pattern.

Start your "position" variable at head

curr = head

This is your finger on the chain

Understand First

Step 1 / 3

Count the Nodes

Same traversal pattern as before, but now you count instead of print.

Start a counter

count = 0

Understand First

Step 1 / 4

Reverse the Chain

This is THE classic linked list interview question. The key insight: you're flipping arrows one by one. But if you flip an arrow without saving where it pointed, you lose the rest of the list forever.

Three pointers: prev=None, curr=head

prev is where we came from. curr is where we are.

prev = None | curr = head

Solve all 3 exercises to unlock completion

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