How do you solve Palindrome Linked List on LeetCode?

Palindrome Linked List (LeetCode #234) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using two pointers helps find the middle efficiently.

What is the brute force solution for Palindrome Linked List?

The brute force approach for Palindrome Linked List is Brute Force, with O(n²) time complexity and O(n) space complexity. The simplest way to check if a linked list is a palindrome is to convert it into a list or string and then check if that list/string reads the same forwards and backwards. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Palindrome Linked List?

Palindrome Linked List uses the following concepts: Linked List, Two Pointers, Stack, Recursion. The recommended patterns to study are: Two Pointers, Linked List.

What are the interview tips for Palindrome Linked List?

Explain your thought process clearly. Mention edge cases during your explanation. Practice coding without looking at references.

What are common mistakes when solving Palindrome Linked List?

Not handling edge cases like single-node lists. Forgetting to restore the original list if required.

#234

Palindrome Linked List

Easy

Linked ListTwo PointersStackRecursionTwo PointersLinked List

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

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Intuition

Time O(n)Space O(1)

To check if a linked list is a palindrome efficiently, we can use the two-pointer technique to find the middle of the list and then reverse the second half to compare it with the first half.

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Algorithm

4 steps

1Step 1: Use two pointers to find the middle of the linked list.
2Step 2: Reverse the second half of the linked list.
3Step 3: Compare the first half and the reversed second half node by node.
4Step 4: Restore the original list (optional) and return the result.

solution.py18 lines

1def isPalindrome(head):
2    slow = fast = head
3    while fast and fast.next:
4        slow = slow.next
5        fast = fast.next.next
6    prev = None
7    while slow:
8        next_node = slow.next
9        slow.next = prev
10        prev = slow
11        slow = next_node
12    left, right = head, prev
13    while right:
14        if left.val != right.val:
15            return False
16        left = left.next
17        right = right.next
18    return True

ℹ

Complexity note: The time complexity is O(n) because we traverse the list a constant number of times. The space complexity is O(1) since we only use a few pointers for traversal and comparison.

1Using two pointers helps find the middle efficiently.
2Reversing the second half allows direct comparison.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.