栈的压入、弹出序列

#include<stdio.h>
#include<string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <stdlib.h>
#include <sstream>
#include <ctype.h>

using namespace std;

class Solution {
public:
	bool IsPopOrder(vector<int> pushV, vector<int> popV) {
		bool isPossible = false;
		int i = 0, j = 0;
		if (popV.size() != pushV.size())
			return false;
		while (j<popV.size())
		{
			/*将pushV中的每个数和popA进行比较，如果不相等则把pushV[i]压入栈，如果相等同时越过
			注：刚开始时s1为空的情况下吧？？*/
			while (i<pushV.size() && pushV[i] != popV[j])
			{
				s1.push(pushV[i]);
				i++;
			}
			cout << s1.size()<<" ";
			cout << i << " ";
			cout << j << endl;
			i++;
			j++;
			/*检查已经入栈的数和下一个出栈数是否相等，如果相等则进行popV下一个数字的比较，同时把s1中相等的数pop出来
			不等则继续进行push的压入操作*/
			while (!s1.empty() && j<popV.size())
			{
				int top = s1.top();
				if (top == popV[j]){
					j++;
					s1.pop();
				}
				else{
					//s1.push(popV[j]);
					//这里只是跳出了第一次循环
					break;
				}
			}
			if (i >= pushV.size() && s1.empty())
				isPossible = true;
		}
		return isPossible;
	}
private:
	stack<int> s1;
};


int main()
{
	Solution s;
	vector<int> pushV = { 1, 2, 3, 4, 5 };
	vector<int>popV = { 4,5, 3, 2, 1 };
	cout << s.IsPopOrder(pushV,popV);
	system("pause");

	return 0;
}