Leetcode 234. Palindrome Linked List

Problem

Given the head of a singly linked list, return true if it is a palindrome or false otherwise.

Algorithm

Use a double-step pointer to find the start of the second half of the linked list (skip if counting), reverse the second half, then compare elements for validation.

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return True

        p1, p2 = head, head
        while p2 and p2.next:
            p1 = p1.next
            p2 = p2.next.next

        if p2:
            p1 = p1.next
        
        p3, p4 = p1, None
        while p3:
            p0 = p3.next
            p3.next = p4
            p4 = p3
            p3 = p0
        
        p1, p2 = head, p4
        while p2:
            if p1.val != p2.val:
                return False
            p1, p2 = p1.next, p2.next

        return True