void K_Reverse( List L, int K ) 
{   //此题已经默认size >= K   因为当size < K时 反转后将不再符合链表的定义
    
    //求出表中元素个数
    int size = 0;
    for (List cur = L->Next; cur != NULL; cur = cur->Next)
        size++;  

    List prv, cur, next, first, head = L;

     //共需要反转 size/K 轮
    for (int i = 1; i <= size / K; i++) 
    {    
        //每轮需要反转K-1次[每两个结点间记作一次]
        int j = K - 1;

        prv = head -> Next;     
        cur = prv-> Next;      
        first = prv;                
               
        while(j--) 
        {
            next = cur -> Next;  
            
            cur -> Next = prv;  
            
            prv = cur;
            cur = next;  
        }
        head -> Next = prv;     
        first -> Next = cur;       
        //下一轮的哨兵位即下一轮的首结点的上一个
        //即上一轮的最后一个结点
        head = first;               
    }
}