234. Palindrome Linked List (Easy)

Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Follow up:
Could you do it in O(n) time and O(1) space?

思路 前一半先reverse，再前后一半顺序比较

如果n为奇数如[1 , 0 , 1] 则中间的数字跳过

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """

        if not head or not head.next:return True
        if not head.next.next and head.val != head.next.val:return False
        elif not head.next.next and head.val == head.next.val:return True
        
        n = 0
        node = head
        while(node):
            n += 1
            node = node.next
        
        node = head.next
        pre = head
        #reverse
        for i in range(1,int(n/2)):
            aft = node.next
            node.next = pre
            pre = node
            node = aft
        if n%2 != 0:node = node.next 
        for i in range(int(n/2)):
            if node.val != pre.val:return False
            pre = pre.next
            node = node.next
        return True