【剑指offer 面试题22】栈的压入、弹出序列

思路：

　　不停地压栈，直到栈头元素与弹出序列的首元素相等则出栈，同时弹出序列后移；若不相等则一直保持压栈，直到压入所有元素后弹出序列仍不为空，则说明无法匹配。

C++:

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

bool IsPreOrder(vector<int>& pushVec, vector<int>& popVec)
{
    if(pushVec.size() != 0 && popVec.size() != 0)
    {
        vector<int>::iterator itPush = pushVec.begin();
        vector<int>::iterator itPop = popVec.begin();

        stack<int> _stack;

        while(itPop != popVec.end())
        {
            while(_stack.empty() || _stack.top() != *itPop)
            {
                if(itPush == pushVec.end())
                    break;

                cout<<"push: "<<*itPush<<endl;
                _stack.push(*itPush);
                itPush++;
            }

            if(_stack.top() != *itPop)
                break;

            cout<<"pop: "<<_stack.top()<<endl;
            _stack.pop();
            itPop++;
        }

        if(_stack.empty() && itPop == popVec.end())
            return true;
    }
    return false;
}

int main()
{
    int apush[] = {1,2,3,4,5};
    int apop1[] = {4,5,3,2,1};
    int apop2[] = {4,3,5,1,2};

    vector<int> vpush(apush, apush + 5);
    vector<int> vpop1(apop1, apop1 + 5);
    vector<int> vpop2(apop2, apop2 + 5);
    cout<<"push: 1 2 3 4 5 pop: 4 5 3 2 1"<<endl;
    cout<<"res = "<<IsPreOrder(vpush, vpop1)<<endl<<endl;
    cout<<"push: 1 2 3 4 5 pop: 4 3 5 1 2"<<endl;
    cout<<"res = "<<IsPreOrder(vpush, vpop2)<<endl;
}

 

测试结果：

push: 1 2 3 4 5 pop: 4 5 3 2 1
push: 1
push: 2
push: 3
push: 4
pop: 4
push: 5
pop: 5
pop: 3
pop: 2
pop: 1
res = 1

push: 1 2 3 4 5 pop: 4 3 5 1 2
push: 1
push: 2
push: 3
push: 4
pop: 4
pop: 3
push: 5
pop: 5
res = 0