第234题：回文链表

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：

1. 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
2. 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
3. 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：

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

class Solution(object):
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if head is None:  # 注意边界
            return True
            
        half_end = self.find_half_end(head)  # 获取前半部分列表的末尾节点
        head_2 = self.half_reverse(half_end.next)  # 将后半链表进行反转
        current_1 = head
        current_2 = head_2
        result = True
        while result and current_2 is not None:
            if current_1.val != current_2.val:
                result = False
            current_1 = current_1.next
            current_2 = current_2.next
        half_end.next = self.half_reverse(head_2)  # 将原链表保持原样
        return result

    def find_half_end(self, head):  # 快慢指针法获取前半部分链表末尾节点
        fast = head
        slow = head
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        return slow

    def half_reverse(self, head):  # 反转链表
        current = head
        prev = None 
        while current is not None:
            next_node = current.next
            current.next = prev
            prev = current
            current = next_node 
        return prev