剑指offer--面试题21

题目：设计包含min函数的栈，pop(),push(),min()的时间复杂度均为O(1)

自己所写代码如下：（写‘栈’的代码还是有些不熟练！）

#include <iostream>

using namespace std;

const int MAX = 100;

class Stack
{
private:
    int values[MAX];
    int topindex;
    int minvalue;
    int minSecvalue;

public:
    Stack();
    virtual ~Stack();

    int top() const;
    void push(int n);
    void pop();
    int min();
    bool empty() const;
};

Stack::Stack()
{
    topindex = 0;
    minvalue = INT_MAX;
    minSecvalue = INT_MAX;
}

Stack::~Stack()
{
}

bool Stack::empty() const
{
    return topindex == 0;
}

int Stack::top() const
{
    int toptemp = topindex;
    if(!empty())
    {
        --toptemp;
        return values[toptemp];
    }
    else
    {
        cerr<<"Stack is empty!"<<endl;
        return INT_MAX;
    }
}

void Stack::push(int n)
{
    if(topindex < MAX)
    {
        values[topindex++] = n;
        if(minvalue > n)
        {
            minSecvalue = minvalue;
            minvalue = n;
        }
    }
    else
        cerr<<"Stack is full!"<<endl;
}

void Stack::pop()
{
    if(!empty())
    {
        topindex--;
        if(values[topindex] == minvalue)
            minvalue = minSecvalue;
    }
    else
        cerr<<"Stack is empty!"<<endl;
}


int Stack::min()
{
    if(!empty())
        return minvalue;
    else
    {
        cerr<<"Stack is empty!"<<endl;
        return INT_MAX;
    }
}

#include "stdafx.h"
#include <iostream>
#include "Stack.h"

using namespace std;

int main()
{
    Stack st;
    for(int i=1; i<=10; i++)
        st.push(i);
    int top = st.top();
    cout<<top<<endl;
    int min = st.min();
    cout<<min<<endl;
    st.pop();
    st.pop();
    top = st.top();
    cout<<top<<endl;
    st.push(2);
    top = st.top();
    cout<<top<<endl;
    min = st.min();
    cout<<min<<endl;

    return 0;
}