grid47 Exploring patterns and algorithms
Oct 17, 2024
5 min read
Solution to LeetCode 206: Reverse Linked List Problem
Given the head of a singly linked list, reverse the list and return the new head of the reversed list. You need to reverse the order of the nodes in the list such that the first node becomes the last, the second becomes the second to last, and so on.
Problem
Approach
Steps
Complexity
Input: The input is the head node of a singly linked list where each node contains an integer value and a reference to the next node.
Example: head = [10, 20, 30, 40]
Constraints:
• 0 <= The number of nodes in the list <= 5000
• -5000 <= Node.val <= 5000
Output: Return the head of the new reversed list.
Example: Output: [40, 30, 20, 10]
Constraints:
• The returned list must reflect the reversed order of the nodes.
Goal: The goal is to reverse the linked list and return the head of the newly reversed list.
Steps:
• Initialize two pointers, one for the current node and one for the previous node (which starts as NULL).
• Iterate through the list while the current node is not NULL.
• For each node, change its next pointer to point to the previous node.
• Move both pointers forward to the next node in the original list.
Goal: The solution should work efficiently for any valid singly linked list within the given constraints.
Steps:
• The list may be empty.
• The list may contain up to 5000 nodes.
Assumptions:
• The linked list is well-formed and the input is always valid.
• Input: Input: head = [1, 2, 3, 4, 5]
• Explanation: In this example, the original list is [1, 2, 3, 4, 5]. After reversing, the new list becomes [5, 4, 3, 2, 1].
• Input: Input: head = [10, 20, 30]
• Explanation: Here, the input list is [10, 20, 30]. After reversing, the result is [30, 20, 10].
• Input: Input: head = []
• Explanation: In this case, the input is an empty list, so the output will also be an empty list.
Approach: The problem can be solved by reversing the pointers in the list either iteratively or recursively.
Observations:
• This is a straightforward problem of manipulating pointers in a singly linked list.
• The iterative approach is generally easier to implement and more space-efficient, as it avoids recursion stack overhead.
Steps:
• Initialize two pointers: one for the previous node (NULL) and one for the current node (head).
• Loop through the list, updating the current node's next pointer to point to the previous node.
• Move the previous pointer to the current node and the current pointer to the next node.
• Repeat until all nodes have been reversed, and then return the previous pointer as the new head of the list.
Empty Inputs:
• If the list is empty (head is NULL), the reversed list is also NULL.
Large Inputs:
• For large lists, ensure the algorithm handles the maximum input size (5000 nodes) efficiently.
Special Values:
• If the list contains a single node, it is its own reversal.
Constraints:
• The solution must work within the time limits for both small and large input sizes.
