Leetcode #2046: Sort Linked List Already Sorted Using Absolute Values
In this guide, we solve Leetcode #2046 Sort Linked List Already Sorted Using Absolute Values 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
Given the head of a singly linked list that is sorted in non-decreasing order using the absolute values of its nodes, return the list sorted in non-decreasing order using the actual values of its nodes. Example 1: Input: head = [0,2,-5,5,10,-10] Output: [-10,-5,0,2,5,10] Explanation: The list sorted in non-descending order using the absolute values of the nodes is [0,2,-5,5,10,-10].
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Linked List, Two Pointers, Sorting
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: head = [0,2,-5,5,10,-10]
Output: [-10,-5,0,2,5,10]
Explanation:
The list sorted in non-descending order using the absolute values of the nodes is [0,2,-5,5,10,-10].
The list sorted in non-descending order using the actual values is [-10,-5,0,2,5,10].
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 sortLinkedList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev, curr = head, head.next
while curr:
if curr.val < 0:
t = curr.next
prev.next = t
curr.next = head
head = curr
curr = t
else:
prev, curr = curr, curr.next
return head
Complexity
The time complexity is , where is the length of the linked list. 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.