Palindrome Linked List

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?

思路：
把链表一分为二。把右边的一半翻转，再逐个比对左右的链表就可以。

/**
 * 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 *p,*pre,*q;
        p=head;
        q=head;
        pre=NULL;
        while(q){
            q=q->next;
            if(!q){
                break;
            }
            pre=p;
            p=p->next;
            q=q->next;
        }
        pre->next=NULL;
        ListNode *r=reverse(p);
        p=head;
        q=r;
        while(p&&q){
            if(p->val!=q->val){
                return false;
            }
            p=p->next;
            q=q->next;
        }
        return true;

    }

    ListNode* reverse(ListNode *head){
        ListNode *h=NULL;
        ListNode *p=head;
        ListNode *tail=NULL;
        while(p){
            ListNode *next=p->next;
            p->next=NULL;
            if(!h){
                h=p;
            }else{
                p->next=h;
                h=p;
            }
            p=next;
        }
        return h;
    }
};