This problem requires reversing a section of a singly linked list between two given positions, left and right. Using pointer manipulation, you must carefully detach and reattach nodes to reverse only the specified segment. The goal is to perform this in-place with minimal extra memory while preserving the rest of the list's structure.
Problem Statement
Given the head of a singly linked list and two integers, left and right, reverse all nodes from the left-th to the right-th position, and return the modified list. The reversal must be done in-place, using pointer manipulation without creating new nodes.
For example, given head = [1,2,3,4,5], left = 2, right = 4, the nodes 2,3,4 should be reversed resulting in [1,4,3,2,5]. If left equals right, the list remains unchanged. Constraints ensure that 1 <= left <= right <= n, with n up to 500 nodes.
Examples
Example 1
Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]
Example details omitted.
Example 2
Input: head = [5], left = 1, right = 1
Output: [5]
Example details omitted.
Constraints
- The number of nodes in the list is n.
- 1 <= n <= 500
- -500 <= Node.val <= 500
- 1 <= left <= right <= n
Solution Approach
Use a dummy node for edge handling
Introduce a dummy node before the head to simplify reversing when left is 1. Move a pointer to the node just before the left-th node to start reconnection after reversal.
Iteratively reverse the sublist
Perform in-place reversal by iterating from left to right, adjusting next pointers for each node. Carefully track previous, current, and next nodes to avoid breaking the list.
Reconnect the reversed segment
After reversing, connect the node before the left-th position to the new head of the reversed sublist, and the tail of the reversed sublist to the node after right. Return dummy.next as the new head.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(n) since each node is visited at most once during traversal and reversal. Space complexity is O(1) because the reversal is done in-place with only a few extra pointers regardless of list size.
What Interviewers Usually Probe
- Checks if the candidate handles edge cases like left == 1 or left == right.
- Looks for in-place pointer manipulation without using extra arrays or stacks.
- Monitors how well the candidate maintains connections before and after the reversed sublist.
Common Pitfalls or Variants
Common pitfalls
- Failing to handle the case where the reversal starts at the head node.
- Incorrectly reconnecting the reversed sublist to the remaining list, breaking links.
- Using extra memory instead of performing true in-place reversal.
Follow-up variants
- Reverse nodes in groups of k instead of a single sublist, extending pointer manipulation skills.
- Reverse a sublist with possible cycles in the linked list, requiring cycle detection.
- Reverse only nodes with even values between left and right, combining conditional logic with pointer manipulation.
How GhostInterview Helps
- Guides step-by-step in detaching, reversing, and reconnecting the sublist pointers correctly.
- Highlights edge cases like left at head or right at tail to prevent pointer mismanagement.
- Optimizes the solution to be in-place, tracking time and space complexity with minimal overhead.
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 best approach to reverse a sublist in 'Reverse Linked List II'?
Use a dummy node, locate the node before left, iteratively reverse nodes from left to right, then reconnect the reversed segment to the rest of the list.
Can I use extra arrays or stacks for this problem?
While possible, optimal solutions use in-place pointer manipulation to meet the problem's space efficiency requirements.
How do I handle the edge case where left equals right?
No reversal is needed; the list remains unchanged, but your code should still correctly return the original head.
What pointers are critical during the reversal?
Track the previous node before the sublist, the current node being reversed, and the next node to maintain proper connections.
How can I verify the reversed segment is correct?
Walk through the list after reversal, checking that nodes between left and right are reversed and all other nodes maintain original order.
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