234. Palindrome Linked List

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

 

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?

 

---

# 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:
        alist=[]
        curr=head
        while curr:
            alist.append(curr.val)
            curr=curr.next
        
        lp, rp=0, len(alist)-1
        while lp<rp:
            if alist[lp]!=alist[rp]:
                return False
            else:
                lp+=1
                rp-=1
        return True