题目来源
Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?
判断单链表是不是回文。最傻的方法就是时间复杂度O(n^2)，或者空间O(n)的方法，这两种就不说了。O(n)时间O(1)空间的做法，怎么做呢？
感觉也是做过类似题目的样子。
就是找出后半段，然后反转一下，再进行比较。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (!head || !head->next)
            return true;
        ListNode *slow = head, *fast = head->next;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next;
            fast = fast->next;
        }
        ListNode *head2 = slow->next;
        head2 = reverseList(head2);
        while (head2) {
            if (head->val != head2->val)
                return false;
            head = head->next;
            head2 = head2->next;
        }
        return true;
    }
    
    ListNode* reverseList(ListNode *node)
    {
        ListNode *next = node->next;
        ListNode *pre = NULL;
        while (node) {
            node->next = pre;
            pre = node;
            node = next;
            if (node)
                next = node->next;
        }
        return pre;
    }
};