Palindrome Linked List

class Solution:
    # @param {ListNode} head
    # @return {boolean}
    def isPalindrome(self, head):
    ## Method 1: space not O(1)
        if not head:
            return True
        n=[]
        while head:
            n.append(head.val)
            head=head.next
        for i in range(len(n)/2+1):
            if n[i]!=n[-i-1]:
                return False
        return True
        
      ## Method 2: two pointers 
      def isPalindrome(self, head):
        if not head or not head.next:
            return True
        # fast to end, slow to middle
        slow,fast=head,head
        while fast and fast.next:
            slow=slow.next
            fast=fast.next.next
        # reverse second half
        p3=slow.next
        slow.next=None
        while p3:# 4 steps of reverse each unit
            p4=p3.next
            p3.next=slow
            slow=p3
            p3=p4
        # conpaire 
        while slow:
            if slow.val!=head.val:
                return False
            slow=slow.next
            head=head.next
        return True