Example 1:
​
Input: 1->2
Output: false
Example 2:
​
Input: 1->2->2->1
Output: true
Follow up:
Could you do it in O(n) time and O(1) space?

/**
 * 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;
        if(!head->next->next) return head->val == head->next->val;
        ListNode* slow=head;
        ListNode* fast=head;
        while(fast->next &&fast->next->next)
        {
            slow=slow->next;
            fast=fast->next->next;
        }
//Reverse List
   //   cout<<slow->val <<endl;
        ListNode* pre=head;
        ListNode* cur=head->next;
        ListNode* tmp=head->next->next;
        head->next=NULL;
        ListNode* new_head=slow->next;
​
        if(fast->next)
        {
             slow=slow->next;
        while(cur != slow)
        {
            tmp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=tmp;
        }
​
        }
        else
        {
        while(cur != slow)
        {
            tmp=cur->next;
            cur->next=pre;
            pre=cur;
            cur=tmp;
        }
​
        }           cur=pre;
​
        cout<<cur->val<<endl;
        while(new_head)
        {
            if(new_head->val !=cur->val ) return false;
            new_head = new_head->next;
            cur=cur->next;
        }
        return true;
​
    }
};