网站改版意义企业建站系统

题目描述：

给你一个单链表的头节点 head ，请你判断该链表是否为

回文链表

。如果是，返回 true ；否则，返回 false 。

 

示例 1：

 

输入：head = [1,2,2,1]
输出：true

示例 2：

 

输入：head = [1,2]
输出：false

 

提示：

• 链表中节点数目在范围[1, 105] 内
• 0 <= Node.val <= 9

 

我的作答：

我的思路是先复制一个一样的链表，再反转这个复制的链表，一个结点一个结点比较，碰到不一样的就return false

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):def isPalindrome(self, head):""":type head: Optional[ListNode]:rtype: bool"""if not head: return Truedef copylist(head): #复制链表dummy = ListNode(0)cur = dummywhile head:cur.next = ListNode(head.val)cur = cur.nexthead = head.nextreturn dummy.next #这个头结点真的好烦def reverse(copy_head): #反转复制的链表cur, pre = copy_head, Nonewhile cur:temp = cur.nextcur.next = prepre = curcur = tempreturn precopy_head = copylist(head)copy_head = reverse(copy_head)cur1, cur2 = head, copy_headwhile cur1 and cur2: #比较if cur1.val!=cur2.val:return Falsecur1 = cur1.nextcur2 = cur2.nextreturn True

缺点是真的很繁琐。。orz

 

参考：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):def isPalindrome(self, head):""":type head: Optional[ListNode]:rtype: bool"""# 保证长度大于1，方便一刀两段if not head.next: return True# 遍历一遍得长度# -------------l = 0cur = headwhile cur:cur = cur.nextl += 1# ---------------# 根据长度反转前面一半的链表# -------------------------pre = Nonecur = headi = 0while l//2 != i:nxt =cur.nextcur.next = prepre = curcur = nxti += 1# -----------------------# 长度为奇数，中间的数不用比较if l % 2 == 1: cur = cur.next# 一一对照即可while cur and pre:if cur.val != pre.val:return Falsecur = cur.nextpre = pre.nextreturn True