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Combining a list with a map

Topic 6 of 6 · Linked lists

What is a linked list?

A linked list stores data in nodes. A node holds a value and a reference to the next node. A reference tells the program which object to follow. The head is the first node; a missing next reference marks the end. To reach a later node, you follow the links from the head.

Example: In 4 → 8 → 15 → None, the node holding 4 points to the node holding 8. Finding 15 requires following two links; there is no direct array-style jump to position 2.

INTERACTIVE WALKTHROUGHPredict · step · explain
SHARED CHAPTER LAB · LINKED

Keep next before changing a link.

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Step 1 of 4 · previous=None; current=1; following=2

Interview reasoningWorked exampleTry it yourself
Self-paced learning
Helpful before this chapter

If these ideas are new, begin with:

Understand the idea

A map locates a node by key; a doubly linked list maintains recency order. Together they support an LRU cache: read/update moves a node to the front, and eviction removes the least-recent tail.

Work through an example

With capacity 2, put A, put B, get A, put C evicts B. The map supplies B's node without searching the list.

Time and space complexity

Expected O(1) get/put with hash lookup and local link changes; O(capacity) storage.

Why it works

Each cached key appears exactly once in both map and recency list; sizes agree.

Common mistakes

  • Updating an existing key must not create a duplicate node.
  • Capacity zero requires an explicit no-storage policy.

Chapter lab · Python implementation

A linked list stores values in nodes connected by references. The lab reverses links while retaining the next unread node. Draw references as arrows: changing a link does not move a node in memory. Ordered traversal is cheap; locating an arbitrary position still requires walking from a known node.

This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.

class Node:
    def __init__(self,value,next=None):
        self.value,self.next=value,next

def reverse(head):
    previous=None
    current=head
    while current is not None:
        following=current.next
        current.next=previous
        previous,current=current,following
    return previous

def values(head):
    output=[]
    while head is not None:
        output.append(head.value)
        head=head.next
    return output

def has_cycle(head):
    slow=fast=head
    while fast is not None and fast.next is not None:
        slow=slow.next
        fast=fast.next.next
        if slow is fast:
            return True
    return False

Run and test the lab

Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.

head=Node(1,Node(2,Node(3)))
assert values(reverse(head))==[3,2,1]
assert reverse(None) is None
assert not has_cycle(Node(1))
a=Node(1); b=Node(2,a); a.next=b; assert has_cycle(a)
CHECK YOUR UNDERSTANDING

Why is a list alone insufficient for fast get?

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