更多“[ ] Prior to engaging in negotiation it is wise to consider one’s own “bottom-line” ”相关问题
  • 第1题:

    在循环双链表的p结点之后插入s结点的操作是______。

    A.p→next=s; p→next→prior=s; S→prior=p; S→next=p→next;

    B.s→next=p; s→next=p→next; p→next=s; p→next→prior=s;

    C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;

    D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=S;


    正确答案:D

  • 第2题:

    在一个双链表中,在*p节点(非尾节点)之后插入一个节点*s的操作是()。

    A.s->prior=p;p->next=s; p->next->prior=s;s->next=p->next;

    B.s->next=p->next;p->next->prior=s;p->next=s;s->prior=p;

    C.p->next=s;s->prior=p;s->next=p->next; p->next->prior=s;

    D.p->prior=s; s->next=p; s->next->prior=p; p->next=s->next;


    O(n)

  • 第3题:

    在循环双链表的p所指向结点之后插入结点s的操作步骤是()。

    A.p->next=s; s->prior=p; p->next->prior=s; s->next=p->prior;

    B.p->next=s; p->prior->next=s;s->prior=p; s->next=p->next;

    C.s->prior=p; s->next=p->next; p->next=s; p->next->prior=s;

    D.s->prior=p; s->next=p->next; p->next->prior=s; p->next =s;


    s->prior=p;s->next=p->next;p->next->prior=s;p->next=s;

  • 第4题:

    在循环双链表的p节点之后插入s节点的操作是______。

    A.p→next=s; p→next→prior=s; s→prior=p; S→next=p→next;

    B.s→next=p; s→next=p→next; p→next=S; p→next→prior=s;

    C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;

    D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=s;


    正确答案:D

  • 第5题:

    设双向循环链表中结点的结构为(data, prior, next)。若想在指针p所指结点之后插入指针s所指结点,则应执行下列哪一个操作?

    A.p->next=s;s->prior=p;p->next->prior=s;s->next=p->next;

    B.s->prior=p;s->next=p->next;p->next->prior=s;p->next=s;

    C.s->prior=p;s->next=p->next;p->next=s;p->next->prior=s;

    D.p->next=s;p->next->prior=s;s->prior=p;s->next=p->next;


    C

  • 第6题:

    【单选题】完成在非空双向循环链表结点p之后插入s的操作是()。

    A.p->next=s ; s->prior=p; p->next->prior=s ; s->next=p->next;

    B.p->next->prior=s; p->next=s; s->prior=p; s->next=p->next;

    C.s->prior=p; s->next=p->next; p->next=s; p->next->prior=s ;

    D.s->next=p->next; p->next->prior=s ; s->prior=p; p->next=s;


    s->next=p->next; p->next->prior=s ; s->prior=p; p->next=s;