链表操作---找到链表的中点

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

ListNode * getmid(ListNode *head)
{
   if(!head||!head->next || !head->next->next)
      return head ;// 空节点，一个节点或者两个节点
  
   ListNode *mid = head ;
   ListNode * fast = head ;
   
   while(NULL!= fast->next && NULL！=fast->next->next)
    {
     mid = mid->next;
     fast = fast->next->next;
   }

    return mid;

}

input:1 2 3 4 5 6 7 8 9 
output:5 6 7 8 9