[Description]
Given a positive integer N representing the number of trains, 0 < N < 10, then enter the train inbound sequence, a total of N trains, each train numbered 1-9 numbers. It is required to sort out the serial number of the outbound train in lexicographic order.  
[Input]
There are several groups of test cases, each group of first line input a positive integer N (0 < N < 10), the second line includes N positive integers, ranging from 1 to 9.  
[Output]
Output the train outbound serial number sorted in lexicographic order, each number separated by spaces, and each output sequence line, specifically sample.  
Sample input 3123
Sample output
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 2 1

[Thoughts on Problem Solving]

Firstly, all possible output sequences are constructed.

Then: output in dictionary order

Initialize a sort vector < string >: Each output is stored in the vector in string form, and then sort the vector in dictionary order.

#include <stack>
#include <iostream>
#include <stack>
#include <algorithm>
#include <vector>

using namespace std;

bool isOutNum(int *push, int *pop, int len)//Judging whether pop is the stack-out sequence of push
{
	if (push == NULL || pop == NULL || len <= 0)
		return false;
	stack<int> Stack;
	int i = 0, j = 0;
	for (i = 0;i<len;i++)//push the numbers on the stack in turn
	{
		Stack.push(push[i]);
		while (j<len && Stack.size() != 0 && pop[j] == Stack.top())//Determine in turn whether each pop sequence value is equal to the top of the stack
		{
			Stack.pop();
			j++;
		}
	}
	return Stack.empty();
}

int main()
{
	int N;
	while (cin >> N)
	{
		//Initialize arrays
        int *pushNum = new int[N];
		int *popNum = new int[N];
     
		for (int i = 0;i<N;i++)
		{
			cin >> pushNum[i];
			popNum[i] = pushNum[i];
		}
        //pop is an ordered arrangement
		sort(popNum, popNum + N);// This is because it's a continuous pointer interval. 

		vector<int> vec(N);//Initialize a vector;
		for (size_t i = 0; i < vec.size(); i++)
		{
			vec[i] = popNum[i]; //To assign values from pop to vec, use a loop
		}

		do
		{
			if (isOutNum(pushNum, popNum, N))//If the arrangement is correct, the output
			{
				for (int i = 0;i<N - 1;i++)
					cout << popNum[i] << " ";
				cout << popNum[N - 1] << endl;
			}
		} while (next_permutation(popNum, popNum + N));//Get the next permutation       
	}
	return 0;
}