c++ - 推送和弹出操作的混合序列为什么这个序列不可能

Question

我正在复习期末考试，但我无法解决这个问题:

Suppose that a client performs an intermixed sequence of stack push and pop operations. The push operations push the integers 0 through 9 in order on to the stack; the pop operations print out the return value. Which of the following sequences could not occur?

(a) 4 3 2 1 0 9 8 7 6 5
(b) 2 1 4 3 6 5 8 7 9 0
(c) 0 4 6 5 3 8 1 7 2 9
(d) 4 6 8 7 5 3 2 9 1 0
(e) All of these sequences are possible

答案是C，但我不确定如何得出这个结论

Answer 1

好的，所以程序总是按 0-9 的顺序压入，所以查看每一行，我们计算出压入和弹出的顺序

**The first line.**   - Stack is
Pushes 0, 1, 2, 3, 4  - [0, 1, 2, 3, 4]
Pops   4, 3, 2, 1, 0  - []
Pushes 5, 6, 7, 8, 9  - [5, 6, 7, 8, 9]
Pops   9, 8, 7, 6, 5  - []

**Second line**  - Stack is  
Pushes 0, 1, 2   - [0, 1, 2]  
Pops   2, 1      - [0]  
Pushes 3, 4      - [0, 3, 4]  
Pops   4, 3      - [0]  
Pushes 5, 6      - [0, 5, 6]  
Pops   6, 5      - [0]  
Pushes 7, 8      - [0, 7, 8]  
Pops   8, 7      - [0]  
Pushes 9         - [0, 9]  
Pops   9, 0      - []  

**Third line**    - Stack is   
Pushes 0          - [0]  
Pops   0          - []  
Pushes 1, 2, 3, 4 - [1, 2, 3, 4]  
Pops   4,         - [1, 2, 3]  
Pushes 5, 6       - [1, 2, 3, 5, 6]  
Pops   6, 5, 3    - [1, 2]  
Pushes 7, 8       - [1, 2, 7, 8]  
Pops   8          - [1, 2, 7]   
Pops   ?            

下一个 pop 必须是 7，因为它是在 8 之前被压入的，它不能是 1。