[Leetcode] 234. Palindrome Linked List

判斷是否是回文

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

先將串列放入list內
判斷第一的位置和最後一個位置的數值相減是否為0，若不為0，則記數+1
判斷記數是否為0，若為0 則True ，反之 False

Python

   # 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: ListNode
   :rtype: bool
   """
   tem=[]
   count = 0
   while(head!= None ):
   tem.append(head.val)
   head=head.next
   length=len(tem)/2 

   if(length==0):
   return True 

   else:
   for i in range(length):
          if(tem[i]-tem[-i-1] != 0):
   count+=1