第234题：回文链表
描述：
请判断一个链表是否为回文链表。（你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？）
示例：
输入: 1->2->2->1
输出: true
解题思路：
- 若想达到O(1) 的空间复杂度，想到对两个列表每个节点依次进行对比
- 需要找到原链表后半部分的节点，将后半部分链表进行反转
- 再对两个链表节点值依次进行对比，从而判断原链表是否为回文链表
Python代码：
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def isPalindrome(self, head): 9 """ 10 :type head: ListNode 11 :rtype: bool 12 """ 13 if head is None: # 注意边界 14 return True 15 16 half_end = self.find_half_end(head) # 获取前半部分列表的末尾节点 17 head_2 = self.half_reverse(half_end.next) # 将后半链表进行反转 18 current_1 = head 19 current_2 = head_2 20 result = True 21 while result and current_2 is not None: 22 if current_1.val != current_2.val: 23 result = False 24 current_1 = current_1.next 25 current_2 = current_2.next 26 half_end.next = self.half_reverse(head_2) # 将原链表保持原样 27 return result 28 29 def find_half_end(self, head): # 快慢指针法获取前半部分链表末尾节点 30 fast = head 31 slow = head 32 while fast.next is not None and fast.next.next is not None: 33 slow = slow.next 34 fast = fast.next.next 35 return slow 36 37 def half_reverse(self, head): # 反转链表 38 current = head 39 prev = None 40 while current is not None: 41 next_node = current.next 42 current.next = prev 43 prev = current 44 current = next_node 45 return prev