89. 回文链表.md

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """

        if not head:
            return False
        if not head.next:
            return True

        # 用于找到中间点的指针
        slow = head
        fast = head
        # 用于反转的指针
        pre = None
        cur = head

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            cur.next = pre
            pre = cur
            cur = slow

        #奇数情况
        if fast:
            slow = slow.next

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