Pop Sequence (25)

输入 1 2 3 4 5
输出 1 4 3 2 5
最大容量3

要输出的第一个数是1
我就从in里取数到stack，直取stack的top元素是1为止
in 2 3 4 5
out 1 4 3 2 5
stack 1
这时同时pop out和stack
in 2 3 4 5
out 4 3 2 5
stack

要输出4
我就从in里取数到stack，直取stack的top元素是4为止
in 5
out 4 3 2 5
stack 2 3 4
这时同时pop out和stack
in 5
out 3 2 5
stack 2 3
发现此时out front和stack top相等
继续pop
in 5
out 5
stack

一直重复，直到out队列为空，也就是说我们可以按out队列的次序输出
最后一步不赘述咯

#include <iostream>
#include <queue>
#include <stack>
using namespace std;
int n;
unsigned stackMaxSize;
bool fun(queue<int> out);
int main()
{
    int k;
    cin >> stackMaxSize >> n >> k;
    queue<int> out;

    for (int j = 0; j < k; j++){
        while (out.empty()!=true){
            out.pop();
        }
        for (int i = 0; i < n; i++){
            int getNum;
            cin >> getNum;
            out.push(getNum);
        }
        if (fun(out)){
            cout << "YES" << endl;
        }
        else{
            cout << "NO" << endl;
        }
    }
}
bool fun(queue<int> out)
{
    queue<int> in;
    for (int i = 1; i <= n; i++){
        in.push(i);
    }

    stack<int> process;
    while (out.empty() != true){
        while (process.empty() || process.top() != out.front()){
            if (in.empty() == true){
                return false;
            }
            process.push(in.front());
            in.pop();
            if (process.size() > stackMaxSize){
                return false;
            }
        }
        process.pop();
        out.pop();
    }
    return true;
}