https://leetcode-cn.com/problems/palindrome-linked-list/
    需要注意的点
    第一种，遍历链表得到数组，复制数组的相反副本，比较数组的值
    第二种，快慢指针

    1. # Definition for singly-linked list.
    2. # class ListNode:
    3. #     def __init__(self, val=0, next=None):
    4. #         self.val = val
    5. #         self.next = next
    6. class Solution:
    7.     def isPalindrome(self, head: ListNode) -> bool:
    9.         if head is None:
    10.             return False
    11.         l = []
    12.         while head is not None:
    13.             l.append(head.val)
    14.             head = head.next
    16.         l1 = l[::-1]
    17.         return l1 == l
    1. # Definition for singly-linked list.
    2. # class ListNode:
    3. #     def __init__(self, val=0, next=None):
    4. #         self.val = val
    5. #         self.next = next
    6. class Solution:
    7.     def isPalindrome(self, head: ListNode) -> bool:
    8.         slow = head
    9.         fast = head
    10.         # 推进快慢指针(节点)
    11.         # 快节点一次两步，慢节点一次一步
    12.         while fast is not None and fast.next is not None:
    13.             slow = slow.next
    14.             fast = fast.next.next
    16.         # 如果快节点不是Null,说明慢节点在中心点
    17.         if fast is not None:
    18.             slow = slow.next
    20.         left = head
    21.         right = self.reverseList(slow)
    23.         while right is not None:
    24.             if left.val != right.val:
    25.                 return False
    27.             left = left.next
    28.             right = right.next
    30.         return  True
    32.         '''
    33.             反转链表
    34.         '''
    35.     def reverseList(self, node: ListNode) -> ListNode:
    36.         pre = None
    37.         cur = node
    38.         while cur is not None:
    39.             next = cur.next
    40.             cur.next = pre
    41.             pre = cur
    42.             cur = next
    44.         return pre