//时间复杂度O(n) 空间复杂度O(1)
    public boolean isPalindrome(ListNode head) {
        ListNode fast = head,slow= head;
        while (fast !=null && fast.next !=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        //如果链表是奇数个结点，把正中的归到左边
        if(fast != null){
            slow = slow.next;
        }
        slow = reverse(slow);
        fast = head;
        
        while (slow != null){
            if(fast.val != slow.val){
                return false;
            }
            fast = fast.next;
            slow = slow.next;
        }
        return true;
        
    }
    
    //反转链表
    public  ListNode reverse(ListNode head){
          ListNode prev = null;
          while (head !=null){
              ListNode next =head.next;
              head.next = prev;
              prev = head;
              head = next;
          }
          return prev;
    }