火车进站(HJ77)

一：解题思路

这道题目比较难，不是数组的全排列，和数组的全排列有点类似，可以放在一起学习。需要仔细阅读题目才能够比较好的理解，其本质就是考察栈数据结构以及对它的后进先出特性的理解。Time:O(n^2),Space:O(n)

二：完整代码示例 (C++版和Java版)

C++代码：

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

using namespace std;

bool isOutNum(vector<int>& push, vector<int>& pop)
{
    if (push.size() == 0 || pop.size() == 0) return false;
    stack<int> stack;
    int j = 0;

    for (int i = 0; i < push.size(); i++)
    {
        stack.push(push[i]);
        while (j < pop.size() && !stack.empty() && pop[j] == stack.top())
        {
            stack.pop();
            j++;
        }
    }

    return stack.empty();
}

int main()
{
    int n = 0;

    while (cin >> n)
    {
        vector<int> pushNum(n,0);
        vector<int> popNum(n,0);

        for (int i = 0; i < n; i++)
        {
            cin >> pushNum[i];
            popNum[i] = pushNum[i];
        }

        sort(popNum.begin(),popNum.end());

        do
        {
            if (isOutNum(pushNum, popNum))
            {
                for (int i = 0; i < popNum.size(); i++)
                {
                    cout << popNum[i] << " ";
                }
                cout << endl;
            }

        } while (next_permutation(popNum.begin(),popNum.end()));
    }

    return 0;
}