Leetcode题目：Palindrome Linked List

题目：

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?

题目解答：题目中要求在O(n)的时间复杂度和O(1)的空间复杂度内解决问题，显然的，复制和使用栈来实现时不可取的。唯一可以想到的思路，就是扭转链表法。

将前半段链表扭转后和后半段链表中的每个节点进行比较，并返回相应结果。

需要注意的是，当链表的节点数奇偶情况不同时，需要有不同的处理细节。

代码如下：

/**
 * 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 == NULL) || (head -> next == NULL))
            return true;
        ListNode * QuickNode = head;
        ListNode * SlowNode = head;
        while((QuickNode != NULL) && (QuickNode -> next != NULL) && (QuickNode -> next -> next != NULL))
        {
            QuickNode = QuickNode -> next -> next;
            SlowNode = SlowNode -> next;
        }
        bool node_num_even = true;
        if(QuickNode -> next == NULL) //奇数个节点
        {
            node_num_even = false;
        }
        ListNode *p = NULL;
        ListNode *q = head;
        ListNode *r = q -> next;
        while(q != SlowNode)
        {
            q -> next = p;
            p = q;
            q = r;
            r = q -> next;
        }
        q -> next = p;
        if(!node_num_even)
        {
            q = q -> next;
        }
        while((q != NULL) && (r != NULL) && (q -> val == r -> val))
        {
            q = q -> next;
            r = r -> next;
        }
        if((q == NULL) && (r == NULL))
            return true;
        else
            return false;
    }
};