南邮OJ 1005 多项式加法（二）

一、首先，这个多项式是一个链表，多项式的每一项是链表一个节点，那么可以想到两种情况：

       1）多项式只有一项或者是多项式的最后一项，那么这个节点就只需要有系数和指数两个元素，且不需要指向下 一个节点。

        2）多项式的其中一项，那么这个节点就需要有系数、指数以及指向下一个节点的指针。

<span style="font-size:18px;"><span style="font-size:18px;">class  Node{private：     int  coef;     int  exp;     Node  *  link;public:     Node (int  c, int  e):coef(c),exp(e)   //多项式不含指向下一个节点的构造函数     {           link = 0;     }     Node (int  c, int  e, Node * next):coef(c),exp(e) //含有指向下一个节点的构造函数     {           link = next;     }}</span></span></span>

另外节点要有插入的功能，所以节点类中要有公有成员函数Insert()函数：

<span style="font-size:18px;"><span style="font-size:18px;">Node  *  Insert(int c, int e){      link = new Node (c,e,link);      return link;}</span></span>

二、链表是由节点构成的，且链表要有添加节点、输出链表以及链表相加的功能。

<span style="font-size:18px;"></pre><pre name="code" class="cpp"><span style="font-size:18px;">class  nodeList{private：      Node * theList;public:     nodeList();     ~nodeList();      void AddNode(istream & in);      void Output(ostream & out);      void ListAdd(nodeList & r);}</span></span>

三、由于对象是类，对于输出操作符不能直接输出节点类，所以我们要做一个运算符重载。

ostream & operator << (ostream & out, const Node & val)

整个思路即为上面的三大块,现在一块一块的分开分析。

第一块，链表元素节点：

<span style="font-size:18px;">class  Node {private:     int coef;   //系数     int exp;    //指数     Node * link;   //指向下一个节点的指针public:    Node (int  c, int  e):coef(c),exp(e)   //多项式不含指向下一个节点的构造函数    {        link = 0;    }    Node (int  c, int  e, Node * next):coef(c),exp(e) //含有指向下一个节点的构造函数    {        link = next;    }    Node * Insert (int c, int e) //在节点后插入一个节点    {        link = new Node(c,e,link);        return link;    }    friend class nodeList;   //链接要操作节点    friend  ostream & operator << (ostream &, const Node &);  //节点的输出用到节点的私有成员}</span>

第二块：链表

<span style="font-size:18px;">class nodeList{    friend ostream & operator << (ostream &, const nodeList & );    friend istream & operator >> (istream &, nodeList &);    friend nodeList & operator + (nodeList &, nodeList &);private:Node * theList;public:nodeList();~nodeList();void AddNode (istream & in);void Output (ostream & out) const;void Listadd (nodeList & r);};nodeList :: nodeList(){theList = new Node (0,-1);theList ->link = theList;}nodeList::~nodeList(){Node * p = theList->link;while (p != theList){theList->link = p->link;delete p;p = theList->link;}delete theList;}void nodeList:: AddNode(istream & in){Node * q = theList;int count = 0;int c,e;for (;;){cin >> c >>e;if (c==0 && e ==-1){break;}else{q = q->Insert(c,e);}count ++;}if (count == 0){q = q->Insert(0,0);cout << 0;}}void nodeList:: Output (ostream & out )const{int i=0,j,k=0;Node * m = theList->link;for (; m!=theList & m->coef == 0;m=m->link)    {        i++;    }    m = theList->link;for (;m!= theList; m=m->link){k++;}m = theList->link;if (k==1&& m->coef == 0){cout << 0 << endl;return;}if (k==i)    {        cout << 0<< endl;        return ;    }    for (j =0; j<i;j++)    {       m=m->link;    }out << *m;m=m->link;for (;m!=theList;m=m->link){if (m->coef > 0){cout << "+";out << *m;}else if (m->coef < 0){out << *m;}}cout << endl;}void nodeList::Listadd(nodeList & r){Node * q,*q1 = theList,*p;p = r.theList->link;q = q1->link;while(p->exp >= 0){   while (p->exp < q->exp){q1=q;q=q->link;}if (p->exp == q->exp){q->coef = p->coef + q->coef;if (p->coef == 0){q1->link = q->link;delete q;q = q1->link;}else{q1=q;q=q->link;}    }elseq1 = q1->Insert(p->coef,p->exp);p=p->link;}}ostream & operator <<(ostream & out, const nodeList & x){x.Output(out);return out;}istream & operator >> (istream & in, nodeList & x){x.AddNode (in);return in;}nodeList & operator + (nodeList & a, nodeList & b ){a.Listadd (b);return a;}</span>

第三块：运算符重载，输出对象是节点类

<span style="font-size:18px;">ostream & operator << (ostream & out, const Node & val){if (val.coef == 1){switch (val.exp){case 0:out << 1;break;case 1:out << "X";break;default : out << "X^" << val.exp; break;}return out;}if (val.coef == -1){switch (val.exp){case 0:out << -1;break;case 1:out << "-X"; break;default : out << "-X^"<< val.exp;break;}return out;}else{out << val.coef;switch (val.exp){case 0:break;case 1:out <<"X"; break;default : out<<"X^" << val.exp;break;}return out;}}</span>