234. 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?

s思路:
1. 套路题。先快慢指针找到中点，然后把后半部分reverse,然后两边都从头开始遍历，看是否相等。

class Solution {public:    bool isPalindrome(ListNode* head) {        //        if(!head) return true;        ListNode* fast=head->next,*slow=head;        //step 1: 找中点        while(fast&&fast->next){            fast=fast->next->next;            slow=slow->next;           }        //step 2:reverse the right half        ListNode* pre=NULL,*pnow=slow->next,*pnext=NULL;        slow->next=NULL;        while(pnow){            pnext=pnow->next;            pnow->next=pre;            pre=pnow;            if(!pnext) break;            pnow=pnext;            }        //step 3:逐一比较        ListNode* l=head,*r=pnow;        while(r){            if(r->val!=l->val) return false;            r=r->next;            l=l->next;        }        return true;    }};