Reorder List
Given a list L: L0 → L1 → … → Ln-1 → Ln, reorder it to: L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …
Output the reordered list as space-separated values.
Example:
1,2,3,4
1 4 2 3
- The input list is split into two parts: the first half and the second half in reverse order.
- The first half of the list is
1, 2and the second half in reverse order is4, 3. - The two halves are then merged in an alternating manner:
1from the first half,4from the second half,2from the first half, and3from the second half. - The final output is the merged list as space-separated values:
1 4 2 3.
Constraints:
- 1 <= number of nodes <= 5 * 10^4
- 1 <= Node.val <= 1000
Background Knowledge
The "Reorder List" problem involves manipulating a linked list, a fundamental data structure in computer science. A linked list is a sequence of nodes, where each node contains a value and a reference (i.e., a "link") to the next node in the list. In this problem, we're given a singly linked list, meaning each node only points to the next node, not the previous one. To solve this problem, you should be familiar with basic linked list operations, such as traversal (iterating through the list) and node manipulation (inserting, deleting, or rearranging nodes).
Understanding the concept of node pointers is crucial. In a linked list, each node has a next pointer that points to the next node in the list. To reorder the list, we'll need to update these pointers to point to different nodes. Additionally, we should be aware of the head and tail of the list, which are the first and last nodes, respectively. In this problem, we're reordering the list in a specific pattern, which requires careful consideration of how to update the node pointers to achieve the desired order.
The problem also involves list reversal, which is a common operation in linked list problems. Reversing a linked list means updating the node pointers so that the list is traversed in the opposite direction. This can be done iteratively or recursively, and it's an essential technique to master when working with linked lists.
Algorithm/Approach
The general approach to solving this problem involves a combination of linked list traversal, node manipulation, and list reversal. We can break down the problem into smaller sub-problems, such as finding the middle of the list, reversing the second half of the list, and then merging the two halves in the desired order. This approach requires careful consideration of how to update the node pointers to achieve the correct order.
Step-by-Step Strategy
To solve this problem, follow these steps:
- Find the middle of the linked list using a slow and fast pointer approach.
- Reverse the second half of the list using an iterative or recursive approach.
- Merge the two halves in the desired order by updating the node pointers.
- Handle edge cases, such as an empty list or a list with only one node.
Common Pitfalls
When implementing the solution, watch out for:
- Null pointer exceptions: Make sure to check for null pointers before accessing or updating node values.
- Infinite loops: Be careful when traversing the list to avoid infinite loops.
- Incorrect pointer updates: Double-check that the node pointers are updated correctly to achieve the desired order.
Time & Space Complexity
The expected time complexity for this problem is O(n), where n is the number of nodes in the linked list, since we need to traverse the list to find the middle and reverse the second half. The space complexity is O(1), as we only need to use a constant amount of space to store the node pointers and other variables.