https://leetcode-cn.com/problems/palindrome-linked-list/ (opens new window) 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

# 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: ListNode) -> bool:

        if head is None:
            return False
        l = []
        while head is not None:
            l.append(head.val)
            head = head.next
        
        l1 = l[::-1]
        return l1 == l

# 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: ListNode) -> bool:
        slow = head
        fast = head
        # 推进快慢指针(节点)
        # 快节点一次两步，慢节点一次一步
        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next

        # 如果快节点不是Null,说明慢节点在中心点
        if fast is not None:
            slow = slow.next
        
        left = head
        right = self.reverseList(slow)

        while right is not None:
            if left.val != right.val:
                return False

            left = left.next
            right = right.next
        
        return  True

        '''
            反转链表
        '''
    def reverseList(self, node: ListNode) -> ListNode:
        pre = None
        cur = node
        while cur is not None:
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next

        return pre