在此对比的值，不是节点
# 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:
dummy_head = ListNode(next=head)
cur = dummy_head.next
val = []
while cur:
val.append(cur.val)
cur = cur.next
return val==val[::-1]使用快慢指针法找到链表的中点，将后半部分链表反转，最后比较前半部分和反转后的后半部分是否相同。
# 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
dummy_head = ListNode(next=head)
slow, fast = dummy_head, dummy_head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
prev, cur = None, slow.next
while cur:
tmp = cur.next
cur.next = prev
prev = cur
cur = tmp
p1, p2 = dummy_head.next, prev
while p2:
if p1.val != p2.val:
return False
p1 = p1.next
p2 = p2.next
return True
