Reverse Nodes in k-Group
Problem Description
Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.
You may not alter the values in the list's nodes, only nodes themselves may be changed.
Example 1:
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
Example 2:
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
Constraints:
- The number of nodes in the list is
n. 1 <= k <= n <= 50000 <= Node.val <= 1000
Follow-up: Can you solve the problem in O(1) extra memory space?
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The most straightforward approach to reverse nodes in k-group involves using an additional data structure, such as a list or stack, to store the k nodes temporarily. Once we have stored k nodes, we reverse them and link them back to the list.
- Traverse the list, extracting groups of k nodes and reverse each group.
- Use a stack or list to reverse the nodes in each group.
- Reconnect the reversed nodes with the rest of the list.
- Continue this process for each k-sized group in the list until you have traversed the entire list.
Code:
- Time Complexity: O(n), where n is the number of nodes in the list.
- Space Complexity: O(k), for storing each group of nodes temporarily.
2. Recursive
Intuition:
The recursive approach is an elegant way to solve the problem. In each recursive call, reverse a k-sized group and link the rest using recursion.
- Base case: If the number of nodes is less than k, directly return the head.
- Recursively reverse the k-sized group, and then link it with the recursively processed next k-sized group.
- Continue doing this for all the nodes in the list.
- Time Complexity: O(n), as each node is traversed once.
- Space Complexity: O(n/k) due to the recursion stack.
3. Iterative
Intuition:
An optimal solution with an iterative approach avoids using extra space apart from the input list itself.
- Use pointers to manage and reverse the k-sized groups in a single pass without additional space for node storage.
- Reverse the nodes within each segment of k-sized nodes by adjusting the pointers directly in the list.
- Utilize a dummy node and handle edge cases to ensure the list is correctly reconstructed after reversal.
Code:
- Time Complexity: O(n), where n is the length of the linked list since each node is processed at most twice.
- Space Complexity: O(1), because the operation is done in place.