class Solution:  def isPalindrome(self, head):    """    判斷一個鏈表是否是回文的，很自然的想法就是兩個指針，一個指針從前往后走，一個指針從后往前走，判斷元素值是否相同，這里要分幾個步驟來進行求解：1、找到鏈表長度的一半，用追趕法，一個指針一次走兩步，一個指針一次走一步2、將后一半數(shù)組轉(zhuǎn)置3、判斷鏈表是否是回文鏈表    :type head: ListNode    :rtype: bool    """    slow = fast = head    while fast and fast.next:      slow = slow.next      fast = fast.next.next     node = None    while slow:      nxt = slow.next      slow.next = node      node = slow      slow = nxt     while node and head:      if node.val != head.val:        return False      node = node.next      head = head.next    return True

# Definition for singly-linked list.# class ListNode:#   def __init__(self, x):#     self.val = x#     self.next = Noneclass Solution:  def isPalindrome(self, head):    """    :type head: ListNode    :rtype: bool    """    res = []    cur = head    while cur:      res.append(cur.val)      cur = cur.next    return res == res[: : -1]