题解 | #判断一个链表是否为回文结构#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 the head
# @return bool布尔型
#
def reverseList(head):  
    prev = None  
    current = head  
    while current:  
        next_node = current.next  # 保存下一个节点  
        current.next = prev  # 反转当前节点的指针  
        prev = current  # 将prev移动到当前节点  
        current = next_node  # 将current移动到下一个节点  
    return prev  
def copyList(head):  
    if not head:  
        return None  
      
    dummy = new_head = ListNode(head.val)  # 创建新链表的虚拟头节点  
    current = head.next  
      
    while current:  
        new_head.next = ListNode(current.val)  # 创建新节点并链接到新链表  
        new_head = new_head.next  
        current = current.next  
      
    return dummy  # 返回新链表的头节点（跳过虚拟头节点）
class Solution:
    def isPail(self, head: ListNode) -> bool:
        # write code here
        if not head or not head.next:
            return True
        node1 = copyList(head)
        node2 = reverseList(head)
        while node1:
            if node1.val == node2.val:
                node1 = node1.next
                node2 = node2.next
            else:
                return False
        return True