DSA · Theory

Construct a linked list from 2D matrix

Theory

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Construct a linked list from 2D matrix

Don’t start Construct a linked list from 2D matrix with jargon. Start with a ticket queue node. That is the whole point.

Construct a linked list from 2D matrix order: meaning → head → next → … → one limit. Dumping ten features helps nobody.

Skip Construct a linked list from 2D matrix and losing the next pointer on delete shows up in a counter.

Use Construct a linked list from 2D matrix when a ticket queue node must stay clear. If a simpler DSA step works, use that instead.

Construct a linked list from 2D matrix trap: losing the next pointer on delete. Dev loses marks for that every viva.

Place Construct a linked list from 2D matrix next to nearby DSA work — head → next → … is the link.

Viva for Construct a linked list from 2D matrix: what it is → head → next → … → the mistake (losing the next pointer on delete).

Diagram

  [10] → [20] → [30] → /
   head              tail

Exam tip

Board: a ticket queue node. Dry-run Construct a linked list from 2D matrix. Name the trap: losing the next pointer on delete.

Example

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

head = Node(1, Node(2, Node(3)))
cur = head
while cur:
    print(cur.val, end=" -> " if cur.next else "\n")
    cur = cur.next

Construct a linked list from 2D matrix — a linked list chains nodes with next pointers instead of contiguous indexes.

Short notes

DefConstruct a linked list from 2D matrix — Dev uses it for a ticket queue node in a counter.
RuleConstruct a linked list from 2D matrix → head → next → ….
RememberConstruct a linked list from 2D matrix + Big-O (a counter).
UseConstruct a linked list from 2D matrix in a counter (a ticket queue node).
TrapConstruct a linked list from 2D matrix — losing the next pointer on delete.
ExConstruct a linked list from 2D matrix → a ticket queue node.

class Node: def __init__(self, val, next=None): self.val = val self.next = next head = Node(1, Node(2, Node(3))) cur = head while cur: print(cur.val, end=" -> " if cur.next else "\n") cur = cur.next

Construct a linked list from 2D matrix

Example

Short notes

Questions

Construct a linked list from 2D matrix

Example

Short notes

Questions