根据存放位置数据的链表P打印链表L的元素

题目：给定一个链表L和另一个链表P，它们包含以升序排列的整数。操作printLots打印L中那些由P所指定的位置上的元素。写出过程printLots（L，P）。只可以使用公有的STL容器操作。该过程的运行时间是多少？

我编写的程序如下：

#include <iostream>

using namespace std;

template <typename Object>
class List
{
    private:
        struct Node                             //List内部维护的节点类
        {
            Object data;
            Node *prev;
            Node *next;

            Node( const Object &d = Object(),Node *p = NULL,Node *n = NULL )
                : data( d ),prev( p ),next( n ) { }
        };

    public:
        class const_iterator                    //常量迭代器
        {
            public:
                const_iterator() : current( NULL )
                { }

                const Object &operator*() const
                { return retrieve(); }

                const_iterator &operator++()
                {
                    current = current->next;
                    return *this;
                }

                const_iterator operator++( int )
                {
                    const_iterator old = *this;
                    ++( *this );
                    return old;
                }

                bool operator== ( const const_iterator &rhs ) const
                { return current == rhs.current; }

                bool operator!= ( const const_iterator &rhs ) const
                { return !( *this == rhs ); }

            protected:
                Node *current;

                Object &retrieve() const
                { return current->data; }

                const_iterator( Node *p ) : current( p )
                { }

                friend class List<Object>;
        };

        class iterator : public const_iterator  //迭代器类继承自常量迭代器类
        {
            public:
            iterator()
            { }

            Object &operator*()
            { return const_iterator::retrieve(); }

            const Object &operator*() const
            { return const_iterator::operator*(); }

            iterator &operator++()
            {
                const_iterator::current = const_iterator::current->next;
                return *this;
            }

            iterator operator++( int )
            {
                iterator old = *this;
                ++( *this );

                return old;
            }

            protected:
                iterator( Node *p ) : const_iterator( p )
                { }

                friend class List<Object>;
        };

    public:
        List()
        { init(); }

        ~List()
        {
            clear();
            delete head;
            delete tail;
        }

        List( const List &rhs )
        {
            init();
            *this = rhs;
        }

        const List &operator= ( const List &rhs )
        {
            if ( this == &rhs )
                return *this;
            clear();
            for ( const_iterator itr = rhs.begin() ; itr != rhs.end() ; ++itr )
                push_back( *itr );

            return *this;
        }

        iterator begin()
        { return iterator( head->next ); }
        const_iterator begin() const
        { return const_iterator( head->next ); }
        iterator end()
        { return iterator( tail ); }
        const_iterator end() const
        { return const_iterator( tail ); }

        int size() const
        { return theSize; }
        bool empty() const
        { return size() == 0; }

        void clear()
        {
            while ( !empty() )
                pop_front();
        }

        Object &front()
        { return *begin(); }
        const Object &front() const
        { return *begin(); }
        Object &back()
        { return *--end(); }
        const Object &back() const
        { return *--end(); }
        void push_front( const Object &x )
        { insert( begin(),x ); }
        void push_back( const Object &x )
        { insert( end(),x ); }
        void pop_front()
        { erase( begin() ); }
        void pop_back()
        { erase( --end() ); }

        iterator insert( iterator itr,const Object &x )
        {
            Node *p = itr.current;
            theSize++;
            return ( p->prev = p->prev->next = new Node( x,p->prev,p ) );
        }

        iterator erase( iterator itr )
        {
            Node *p = itr.current;
            iterator retVal( p->next );
            p->prev->next = p->next;
            p->next->prev = p->prev;
            delete p;
            theSize--;

            return retVal;
        }

        iterator erase( iterator start,iterator end )
        {
            for ( iterator itr = start ; itr != end; )
                itr = erase( itr );

            return end;
        }

    private:
        int theSize;
        Node *head;
        Node *tail;

        void init()
        {
            theSize = 0;
            head = new Node;
            tail = new Node;
            head->next = tail;
            tail->prev = head;
        }
};

void printLots( List<int> &L,List<int> &P )         //打印出L链表 指定位置（P中存放）的值
{
    List<int>::const_iterator itr = L.begin();
    int count = 0;                                  //P是升序排列那么不必每次遍历L，维护计数器一直递增迭代器即可

    for ( List<int>::const_iterator itr_P = P.begin() ; itr_P != P.end() ; ++itr_P ){
        int pos = *itr_P;
        while ( pos != count ){
            if ( itr == L.end() )
                return ;
            count++;
            itr++;
        }
        if ( itr != L.end() )
            cout << *itr << endl;
    }

    return ;
}

int main( void )
{
    List<int> p,temp;
    p.push_back( 0 );
    p.push_back( 5 );

    temp.push_back( 43 );
    temp.push_back( 4353 );

    cout << "The pos List:" << endl;

    for ( List<int>::iterator iter = p.begin() ; iter != p.end() ; ++iter )
        cout << *iter << " ";

    cout << endl;

    cout << "The temp List:" << endl;
    printLots( temp,p );

    return 0;
}

运行结果：

printLots过程中temp链表迭代P中的最大值 次，所以运行时间是取决于P中最后一个元素N，O(N)。