Leetcode#234. 回文链表

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解题思路：

• 快慢指针找到链表中点；
• 反转链表中点->next，即链表的后半部分；
• 同时遍历判断反转后的后半部分与前半部分的值的大小；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:

    ListNode* reverse(ListNode* head)
    {
        ListNode* p = NULL;
        ListNode* q = NULL;

        while(head != NULL)
        {
            q = head->next;
            head->next = p;
            p = head;
            head = q;
        }
        return p;
    }
    bool isPalindrome(ListNode* head) 
    {
        if(head == NULL || head->next == NULL)
            return true;
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast->next && fast->next->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }

        slow->next = reverse(slow->next);
        slow = slow->next;

        while(slow != NULL)
        {
            if(head->val != slow->val)
                return false;
            slow = slow->next;
            head = head->next;
        }
        return true;
    }
};