Palindrome Linked List

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

 

Example 1:

Input: head = [1,2,2,1]
Output: true

Example 2:

Input: head = [1,2]
Output: false

 

Constraints:

  • The number of nodes in the list is in the range [1, 105].
  • 0 <= Node.val <= 9

 

Follow up: Could you do it in O(n) time and O(1) space?SOLUTION:
# 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:
        curr = head
        stack = []
        while curr:
            stack.append(curr.val)
            curr = curr.next
        curr = head
        while curr:
            if curr.val != stack.pop():
                return False
            curr = curr.next
        return True

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