5 - Easy - 回文链表 - （回文链表，快慢指针）

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false
示例 2:

输入: 1->2->2->1
输出: true
进阶：
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

O(n) time and O(n) space

class Solution:
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        stack = []
        while head != None:
            stack.append(head.val)
            head = head.next
        return stack == stack[::-1]

O(n) time and O(1) space 快慢指针，找到中点，翻转后半段，然后两个半个链表判断是否回文

class Solution:
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head or not head.next:
            return True
        slow = fast = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next

        slow = slow.next
        slow = self.reverseList(slow)

        while slow:
            if head.val != slow.val:
                return False
            slow = slow.next
            head = head.next
        return True

    def reverseList(self, head):
        new_head = None
        while head:
            p = head
            head = head.next
            p.next = new_head
            new_head = p
        return new_head