Python3实现的判断回文链表算法示例

class Solution:
  def isPalindrome(self, head):
    """
    判断一个链表是否是回文的，很自然的想法就是两个指针，一个指针从前往后走，一个指针从后往前走，判断元素值是否相同，这里要分几个步骤来进行求解：
1、找到链表长度的一半，用追赶法，一个指针一次走两步，一个指针一次走一步
2、将后一半数组转置
3、判断链表是否是回文链表
    :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 = None
class 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]