FDS作业记录 · 栈和队列

1、

Push 5 characters ooops onto a stack. In how many different ways that we can pop these characters and still obtain ooops?

A.1

B.3

C.5

D.6

stack是一种FILO ( First In Last Out ) 结构，故p和s两个字母必须在入栈后立即出栈，故不同的情况仅发生在前三个o上。

分析可知必须是3个in，3个out，且in的数量不得少于out，故有：

① in in in out out out

② in in out out in out

③ in in out in out out

④ in out in in out out

⑤ in out in out in out

共五种情况。

2、

Represent a queue by a singly linked list. Given the current status of the linked list as 1->2->3 where x->y means y is linked after x. Now if 4 is enqueued and then a dequeue is done, the resulting status must be:

A.1->2->3

B.2->3->4

C.4->1->2

D.the solution is not unique

queue是FIFO ( First In First Out ) 结构，故enqueue了4，dequeue了1，只剩下2->3->4

3、

Suppose that an array of size 6 is used to store a circular queue, and the values of front and rear are 0 and 4, respectively. Now after 2 dequeues and 2 enqueues, what will the values of front and rear be?

A.2 and 0

B.2 and 2

C.2 and 4

D.2 and 6

两次入队两次出队，入队一次rear+1，出队一次front+1

0 + 2 = 2；

4 + 2 = 6，6 % 6 = 0；

故front和rear最终值分别为2，0

4、

Suppose that all the integer operands are stored in the stack S1​, and all the operators in the other stack S2​. The function F() does the following operations sequentially:

• (1) Pop two operands a and b from S1​;
• (2) Pop one operator op from S2​;
• (3) Calculate b op a; and
• (4) Push the result back to S1​.

Now given { 5, 8, 3, 2 } in S1​ (where 2 is at the top), and { *, -, + } in S2​ (where + is at the top). What is remained at the top of S1​ after F() is executed 3 times?

A.-15

B.15

C.-20

D.20

根据题述步骤：

① 从S1中pop两个数：2，3

② 从S2中pop一个运算符：+

③ 计算 3 + 2 = 5

④ push 5 到S1中，S1 = {5，5，8}

……

以此类推，5 - 8 = -3，5 * (-3) = -15，故选A