向列表框list的0行添加字符串变量text内容的代码为list.insert(0,text)

Which operations of linked list are more efficient than array based list? A: insert B: search C: remove D: setPos

Given an empty linked list, the cost to insert n elements in the list is ___? A: Θ(1) B: Θ(n) C: Θ(n*n) D: Θ(log n)

若有List=[2,5,1,7],执行语句List.insert(3,9),List中的值为_______。() A: [9, 2, 5, 1, 7] B: [2, 5, 1, 9, 7] C: [2, 5, 1, 7, 9] D: [2, 5, 1, 7, 3, 9]

已知不带头结点的单链表L，下面用函数实现的在第一个元素前面插入值为x的元素结点的算法错误的是（ ） A: void insert(List *L,elemtype x){ node d=new node(x); d.next=*L; *L=&d;} B: List * insert(List *L,elemtype x){ node d=new node(x); d.next=*L; *L=&d; return L;} C: void insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&d; p->next=q; L=&q;} D: List insert(List *L,elemtype x){ node d=new node(x); d.next=*L; return &d;}

6. The most common operation for setting a linked list is to insert a node at the end and delete the tail node, so () is the most time-saving option. A: doubly linked circular list with a head node B: singly linked list C: singly linked circular list with a tail pointer D: singly linked circular list

通过( )函数将元组或字符串转化成列表 A: list() B: append() C: copy() D: insert()

You<br/>are given pointers to first and last nodes of a singly linked list,<br/>which of the following operations are dependent on the length of the<br/>linked list? () A: Delete<br/>the first element B: Insert<br/>a new element as a first element C: Delete<br/>the last element of the list D: Add<br/>a new element at the end of the list

已知不带头结点的单链表L，下面用函数实现的在第一个元素前面插入值为x的元素结点的算法错误的是（ ） A: void insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&d; q->next=p; L=&q;} B: List * insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&d; q->next=p; L=&q; return L;} C: void insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&d; p->next=q; L=&q;} D: List * insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&d; q->next=p; return &q;}

已知不带头结点的单链表L，下面用函数实现的在第一个元素前面插入值为x的元素结点的算法错误的是（ ） A: void insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&amp;d; q-&gt;next=p; L=&amp;q;} B: List * insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&amp;d; q-&gt;next=p; L=&amp;q; return L;} C: void insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&amp;d; p-&gt;next=q; L=&amp;q;} D: List * insert(List *L,elemtype x){ ptr p=*L; node d=new node(x); ptr q=&amp;d; q-&gt;next=p; return &amp;q;}