在一个双链表中,在* $p$ 结点之后插入一个结点*s 的操作是[input=type:blank,size:4][/input]。
A: s-> prior =p ; p->n e x t=s ; p-> next -> prior =s ; s->n e x t=p-> next s -> next = p -> next ；
B: p -> next -> prior = s ; p -> next = s ; s \rightarrow> prior = p
C: p -> next = s ; s \rightarrow> prior = p ; s -> next = p -> next ; p -> next -> prior = s ;
D: p -> next -> prior = s ; s -> next = p -> next ; s \rightarrow prior = p ; p -> next = s
A: s-> prior =p ; p->n e x t=s ; p-> next -> prior =s ; s->n e x t=p-> next s -> next = p -> next ；
B: p -> next -> prior = s ; p -> next = s ; s \rightarrow> prior = p
C: p -> next = s ; s \rightarrow> prior = p ; s -> next = p -> next ; p -> next -> prior = s ;
D: p -> next -> prior = s ; s -> next = p -> next ; s \rightarrow prior = p ; p -> next = s
举一反三
- 完成在双循环链表结点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()->()prior=p;()s()->()next=p()->()next;()p()->()next()->()prior=s();()p()->()next=s;
- 在循环双链表的p所指结点后插入s所指结点的操作是（ ）。 A: s-＞prior=p; s-＞next=p-＞next; p-＞next=s; p-＞next-＞prior=s; B: p-＞next=s; p-＞next-＞prior=s; s-＞prior=p; s-＞next=p-＞next; 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;
- 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； B: p—＞next=s；s—＞prior=p；p—＞next—＞prior=s；p—＞next—＞prior=s；s—＞prior=p；s—＞next=p—＞next；s—＞next=p—＞next C: s—＞prior=p； D: s—＞prior=p；s—＞next=p—＞next；s—＞next=p—＞next；p—＞next=s；p—＞next—＞prior=s；p—＞next—＞prior=s；p—＞next=s；
- 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；
- 在循环双链表的P所指节点之后插入s所指节点的操作是( ) A: p->next=s; s->prior=p; p->next->prior=s; s->next=p->next; B: p->next=s; p->next->prior=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=s;p->next->prior=s;