OFFER prove safety of two stacks queue (nine degrees OJ1512)

2019 Unicorn Enterprises heavily recruiting engineers Python standard >>>

Subject description:

Two stacks to achieve a queue, the completion queue Push and Pop operations.
Queue elements int.

 

Input:

Each input file contains a test sample.
For each test sample, a first line of the input n (1 <= n <= 100000), the number representing the operation of the queue.
The next n lines, each line of the input queue operation:
1. Push the PUSH a X-integer x (x> = 0) to the queue
2. POP pop a number from the queue.

 

Output:

Corresponding to each test case, all the printing operations of the digital pop pop from the queue. If the pop operation performed, the queue is empty, the print -1

Sample input:

3
PUSH 10
POP
POP

Sample output:

10
-1

 

Problem-solving ideas:

First subject of the request queue represents two stacks. It is best to use two stacks myself, although I do not know how OJ is detected, it should use a queue is not enough.

Then we remembered the Tower of Hanoi, the stack itself is the last in the queue is FIFO. So how to design it?

Two stacks, one stack for the first PUSH, a second stack for POP.

Issues to note is that the order of elements around the middle of the two stacks can not be arbitrary. As long as there are elements in the stack 2 is present, it can not stack a stack 2 to the pressure element . such as

New elements stack 1 3

2 is a stack of elements 21,

Then the two stack list is:

3
2 1

To order this time out of the stack must be

1 2 3

At this time, if the stack 2 onto the stack 1, the order of the stack becomes

3 1 2

Thus, it should not be guaranteed in the stack 2 when the element is empty, stack 2 again pushed onto the stack of element 1 .

Code:

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <string.h>
#define MAX 100001
int stack1[MAX],stack2[MAX],top1,top2;
int main(void){
    int n,i,m;
    char input1[10];
    while(scanf("%d",&n)!=EOF && n<=100000 && n>= 1){
        top1 = top2 = 0;
        memset(&stack1,0,sizeof(int)*MAX);
        memset(&stack2,0,sizeof(int)*MAX);
        for(i=0;i<n;i++){
            scanf("%s",input1);
            if(strcmp(input1,"PUSH") == 0){
                scanf("%d",&m);
                stack1[top1++] = m;
            }else{
                if(top2 == 0){//判断栈2是否还存在元素
                    while(top1){
                        stack2[top2++] = stack1[--top1];
                    }
                }
                if(top2)//如果栈2中有元素，则弹出，否则，表示两个栈都没有元素
                    printf("%d\n",stack2[--top2]);
                else{
                    printf("-1\n");
                }
            }
        }
    }
    return 0;
}

 

Reproduced in: https: //my.oschina.net/u/204616/blog/545514