参考答案和解析
参考答案:C
更多“—What time is the next flight to Washington? —_______A.It’s a smaller one.B.It’s too la ”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:

    完成在双循环链表结点p之后插入s的操作为: The operation to insert s after the doubly circular linked list’s node p is: (There are more than one answers.)

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

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

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

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


    BC

  • 第3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;

  • 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:

    在循环双链表的p所指结点之后插入s所指结点的操作是()。

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

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

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

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


    s->left=p; s->right=p->right; p->right=s; p->right->left=s;

  • 第6题:

    在循环双链表的p所指结点之后插入s结点的操作是()

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

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

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

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


    s->left=p; s->right=p->right; p->right=s; p->right->left=s;