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
#
class Solution:
    def isPail(self , head: ListNode) -> bool:
        # write code here
        cur=head
        a=[]
        while cur:
            a.append(cur.val)
            cur=cur.next
        r=len(a)-1
        l=0
        while l<r:
            if a[l]!=a[r]:
                return False
            l+=1
            r-=1
        return True

class Solution:
    def isPail(self , head: ListNode) -> bool:
        # write code here
        def LenthList(head):    # 计算链表长度
            count = 0
            while head:
                count += 1
                head = head.next
            return count
        
        def ConList(head):  # 逆转链表，方便对比
            p = ListNode(0)
            while head:
                q = head
                head = q.next
                q.next = p.next
                p.next = q
            return p.next

        len = LenthList(head)
        front = p = ListNode(0)
        behind = ListNode(0)
        count = 0
        while head: # 将链表一分为二，前半段和后半段
            if len // 2 == count:
                behind.next = head
                break
            p.next = ListNode(head.val)
            head = head.next
            p = p.next
            count += 1
        Conbehind = ConList(behind.next)
        front = front.next
        while front and Conbehind:  # 对比两个链表，不相等则False
            if front.val != Conbehind.val:
                return False
            front = front.next
            Conbehind = Conbehind.next
        return True

class Solution:
    def isPail(self , head: ListNode) -> bool:
        # write code here
        list1 = []
        while head != None:
            list1.append(head.val)
            head = head.next
        print(list1)
        left = 0
        right = len(list1)-1
        while left < right:
            if list1[left] != list1[right]:
                return False
            left += 1
            right -= 1
        return True

class Solution:
    def isPail(self , head ):
        # write code here
        alist = []
        while(head!=None):
            alist.append(head)
            head = head.next
        n = len(alist)
        if n==0:
            return False
        for i in range((n-1)/2+1):
            if alist[i].val!=alist[n-i-1].val:
            
                return False
        return True 

class Solution:
    def isPail(self , head: ListNode) -> bool:
        res = []
        cur = head 
        while cur:
            res.append(cur.val)
            cur = cur.next
        return res == res[::-1]