面试常见算法习题2

/*2010/10/18 July2.设计包含min函数的栈。定义栈的数据结构，要求添加一个min函数，能够得到栈的最小元素。要求函数min、push以及pop的时间复杂度都是O(1)。*/class Stack{private:    struct Link    {        int data;        Link* next;        Link(const int& dat, Link* nxt):data(dat), next(nxt){}        ~Link(){}    } *pHead;    Link *pMinimum;    public:        Stack():pHead(NULL), pMinimum(NULL){}        Stack(const int& initial)        {            pHead = new Link(initial, NULL);            pMinimum = new Link(initial, NULL);        }        ~Stack()        {            Link* tmpNode = NULL;            while(pHead != NULL)            {                tmpNode = pHead;                pHead = pHead->next;                delete tmpNode;                tmpNode = NULL;            }            while(pMinimum != NULL)            {                tmpNode = pMinimum;                pMinimum = pMinimum->next;                if(tmpNode != NULL)                {                    delete tmpNode;                    tmpNode = NULL;                }            }        }        bool push(const int& newData)        {            pHead = new Link(newData, pHead);            if(pMinimum == NULL || pMinimum->data >= newData)            {                pMinimum = new Link(newData, pMinimum);            }            return true;        }        bool pop(int& iData)        {            if(pHead == NULL)            {                return false;            }            iData = pHead->data;            Link* tmpNode = pHead;            pHead = pHead->next;            delete tmpNode;                        if(iData == pMinimum->data)            {                tmpNode = pMinimum;                pMinimum = pMinimum->next;                delete tmpNode;            }            return true;        }        bool min(int& iMinimum) const        {            if(pMinimum == NULL)            {                return false;            }            iMinimum = pMinimum->data;            return true;        }        bool empty() const        {            return pHead == NULL;        }};