234. 回文链表
- 回文链表
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
思路
找到链表的中间节点 ，求链表长度，若为奇数，中间节点再往后移，在将中间节点以后一段链表翻转，最后遍历链表
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
#z找到中间节点
def middle(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
#求链表长度
def length(self, head: Optional[ListNode]) -> int:
n = 0
cur = head
while cur:
n += 1
cur = cur.next
return n
#将链表翻转
def reverse(self, head: Optional[ListNode]) -> Optional[ListNode]:
pre = None
cur = head
while cur:
nxt = cur.next
cur.next = pre
pre = cur
cur = nxt
return pre
def isPalindrome(self, head: Optional[ListNode]) -> bool:
n = self.length(head)
head2 = self.middle(head)
#如果链表长度为奇数 则head2往后移
if(n & 1):
head2 = head2.next
head2 = self.reverse(head2)
while head2:
if head.val != head2.val:
return False
head = head.next
head2 = head2.next
return True
