Leetcode #2816: Double a Number Represented as a Linked List
In this guide, we solve Leetcode #2816 Double a Number Represented as a Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given the head of a non-empty linked list representing a non-negative integer without leading zeroes. Return the head of the linked list after doubling it.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Linked List, Math
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input: head = [1,8,9]
Output: [3,7,8]
Explanation: The figure above corresponds to the given linked list which represents the number 189. Hence, the returned linked list represents the number 189 * 2 = 378.
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def doubleIt(self, head: Optional[ListNode]) -> Optional[ListNode]:
def reverse(head):
dummy = ListNode()
cur = head
while cur:
next = cur.next
cur.next = dummy.next
dummy.next = cur
cur = next
return dummy.next
head = reverse(head)
dummy = cur = ListNode()
mul, carry = 2, 0
while head:
x = head.val * mul + carry
carry = x // 10
cur.next = ListNode(x % 10)
cur = cur.next
head = head.next
if carry:
cur.next = ListNode(carry)
return reverse(dummy.next)
Complexity
The time complexity is O(n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.