This problem is solved by applying merge sort directly on the linked list, carefully handling pointers to split and merge sublists. The key is to maintain O(n log n) time complexity while using minimal extra space. GhostInterview guides you through breaking the list, merging sorted halves, and handling edge cases for empty or single-node lists.
Problem Statement
Given the head of a singly linked list, implement a function that returns the list sorted in ascending order. You must manipulate the node pointers without converting the list into another data structure, ensuring efficiency for large lists.
For example, given head = [4,2,1,3], your function should return [1,2,3,4]. Given head = [-1,5,3,4,0], it should return [-1,0,3,4,5]. The solution should handle lists with up to 50,000 nodes and node values in the range [-105, 105].
Examples
Example 1
Input: head = [4,2,1,3]
Output: [1,2,3,4]
Example details omitted.
Example 2
Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
Example details omitted.
Example 3
Input: head = []
Output: []
Example details omitted.
Constraints
- The number of nodes in the list is in the range [0, 5 * 104].
- -105 <= Node.val <= 105
Solution Approach
Use Fast and Slow Pointers to Split the List
Apply the two-pointer technique to locate the middle of the linked list, splitting it into two halves. This ensures each recursive merge operates on roughly equal-sized sublists, maintaining O(n log n) time complexity.
Recursive Merge Sort on Sublists
Recursively sort each half of the list by calling the sort function on the left and right segments. This step relies on pointer manipulation instead of array indices and avoids extra array allocation, preserving linked-list efficiency.
Merge Two Sorted Sublists Efficiently
Iteratively or recursively merge the two sorted halves by adjusting the next pointers. Carefully track the head and tail of the merged list to prevent cycles and preserve the ascending order, which is critical in linked-list pointer-based sorting.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(n log n) due to the divide-and-conquer merge sort approach. Space complexity is O(log n) for recursion stack; no additional arrays are used, keeping pointer manipulation efficient.
What Interviewers Usually Probe
- The interviewer may hint at avoiding array conversion for true pointer manipulation.
- Expect discussion on fast and slow pointers for splitting the linked list.
- They often ask how to merge sublists without extra space to verify understanding of linked-list operations.
Common Pitfalls or Variants
Common pitfalls
- Accidentally creating cycles when reconnecting nodes during merge.
- Not handling empty lists or single-node lists as base cases, causing null pointer errors.
- Failing to update head or tail pointers correctly, leading to lost nodes in the final sorted list.
Follow-up variants
- Sort a doubly linked list using similar divide-and-conquer strategies but with backward pointers.
- Sort a linked list where nodes contain extra data that must remain paired with values.
- Implement iterative bottom-up merge sort instead of recursive sorting to reduce stack usage.
How GhostInterview Helps
- Breaks down the recursive merge steps with exact pointer manipulations for clarity.
- Highlights edge cases like empty or single-node lists to prevent runtime errors.
- Provides interactive guidance on merging sublists to maintain ascending order efficiently.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the most efficient way to sort a linked list in LeetCode 148?
Using merge sort directly on the linked list with careful pointer manipulation ensures O(n log n) time and minimal extra space.
Can I convert the linked list to an array and sort it?
While possible, this approach loses the pointer manipulation pattern and increases space complexity, which may be penalized in interviews.
How do fast and slow pointers help in Sort List?
They locate the middle node efficiently to split the list into halves for recursive merge sort, maintaining balanced recursion.
What are common mistakes when merging sorted linked lists?
Forgetting to update next pointers, losing track of head and tail, or creating cycles are frequent pitfalls that break the sorted list.
Does the solution work for very large lists?
Yes, the O(n log n) merge sort with pointer manipulation handles up to 50,000 nodes efficiently without extra arrays.
Need direct help with Sort List instead of spending more time grinding it?
Download GhostInterview when you want a LeetCode solver, not another long practice loop. Capture Sort List from a screenshot, get the answer path and complexity, and use supported stealth workflows that stay outside captured layers.
Capture the prompt fast instead of rewriting the problem by hand.
Get the solution path, trade-offs, and complexity summary in one pass.
Stay outside captured layers on supported screen-share workflows.
Stay in the same pattern family
Reorder List requires careful pointer manipulation in a singly linked list to interleave nodes from the ends without altering values.
Open problem page#142 Linked List Cycle IIIdentify the start of a cycle in a linked list using pointer manipulation, efficiently handling edge cases without modifying the list.
Open problem page#141 Linked List CycleDetermine if a given linked list contains a cycle using pointer manipulation or hashing, focusing on detecting repeated nodes efficiently.
Open problem page