Palindrome Linked List
Given a linked list (as comma-separated values), return True if it is a palindrome.
Example:
1,2,2,1
True
- The input linked list is
1,2,2,1, which can be visualized as a sequence of nodes with values 1→2→2→1. - To check if it's a palindrome, we compare the first and last nodes, then the second and second-to-last nodes, and so on.
- Since the values are symmetric (1=1 and 2=2), the linked list is a palindrome.
- The function returns
Truebecause the input linked list reads the same backward as forward.
Constraints:
- 1 <= number of nodes <= 10^5
- 0 <= Node.val <= 9
Background Knowledge
A linked list is a linear data structure where each element is a separate object, and each element (or "node") points to the next node in the sequence. This structure allows for efficient insertion and deletion of nodes at any point in the list. In the context of this problem, we're dealing with a singly linked list, where each node only points to the next node.
To determine if a linked list is a palindrome, we need to compare the first half of the list with the reversed second half. A palindrome is a sequence that reads the same backward as forward. In the case of a linked list, this means that the values of the nodes should be the same when traversed from the beginning and from the end. Understanding the concept of a palindrome and how to reverse a linked list (or at least, how to access its elements in reverse order) is crucial for solving this problem.
The key concepts to grasp here include pointer manipulation (to traverse the linked list), reversing a linked list (or simulating its reversal), and comparing elements for equality. These concepts are fundamental in solving linked list problems and are essential for understanding the algorithmic approach to determining if a linked list is a palindrome.
Algorithm/Approach
The general approach to solving this type of problem involves a combination of two-pointer techniques and linked list reversal. The two-pointer technique can be used to find the middle of the linked list, and then one of the pointers can be used to reverse the second half of the list. Alternatively, a stack can be used to store the first half of the list, allowing for a straightforward comparison with the second half. The choice of algorithm depends on the desired time and space complexity.
Step-by-Step Strategy
- Identify the Middle: Find the middle of the linked list. This can be done using the two-pointer technique, where one pointer moves twice as fast as the other.
- Reverse the Second Half: Reverse the second half of the linked list. This step is crucial for comparing the first and second halves.
- Compare Elements: Compare the elements of the first half and the reversed second half. If all elements match, the list is a palindrome.
- Restore the List (Optional): If the original list needs to be preserved, restore it after the comparison.
Common Pitfalls
- Incorrectly Identifying the Middle: Failing to correctly identify the middle of the list can lead to incorrect results.
- Not Reversing Correctly: The second half of the list must be properly reversed for the comparison to work.
- Not Handling Edge Cases: Failing to handle edge cases, such as an empty list or a list with a single node, can lead to errors.
Time & Space Complexity
The expected time complexity for this problem is O(n), where n is the number of nodes in the linked list, because we need to traverse the list at least once. The space complexity can vary depending on the approach: using a stack to store the first half of the list results in a space complexity of O(n), while reversing the second half of the list in-place can achieve a space complexity of O(1).