实现两个指数递减多项式的和与积
程序员文章站
2022-03-06 16:52:45
...
有两个指数递减的一元多项式,写一程序先求这两个多项式的和,再求它们的积。
【提示】 用带表头结点的单链表作为多项式的存储表示;要建立两个单链表;多项式相加就是要把一个单链表中的结点插入到另一个单链表中去,要注意插入、删除操作中指针的正确修改。
#include#include using namespace std; /** 数据结构习题1 多项式的相加和相乘 @刘辉 **/ struct Node{ int data; int index; Node* next; }; Node *insertList(Node* head,Node* p); //插入链表 Node *createList(); //创建链表 void printList(Node *head); //打印链表 Node *addList(Node *p1,Node *p2); //实现加法运算 Node *createList() { int index,data; Node *p,*head,*q; head = new Node; p = head; cout>index; cout>data; while(index!=0) { q = new Node; q->index = index; q->data = data; p->next = q; p = q; cout>index; cout>data; } p->next = NULL; return head; } //多项式相加 Node *addList(Node *p1,Node *p2) { int add; Node *temp,*head,*p3; p1 = p1->next; p2 = p2->next; head = temp = new Node; head->next = NULL; while(p1&&p2) { if(p1->index==p2->index) { add = p2->data + p1->data; if(add!=0) { p3 = new Node; p3->index = p2->index; p3->data = add; p3->next = NULL; } p1 = p1->next; p2 = p2->next; } else if(p1->index index) { p3 = new Node; p3->data = p2->data; p3->index = p2->index; p3->next = NULL; p2 = p2->next; } else { p3 = new Node; p3->data = p1->data; p3->index = p1->index; p3->next = NULL; p1 = p1->next; } if(head->next ==NULL) { head->next = p3; temp = p3; } else { temp->next = p3; temp = p3; } } temp->next = p1?p1:p2; return head; } //多项式相乘 Node* mulList(Node *p1,Node *p2) { Node *head,*temp,*s,*r,*q; head = new Node; head->next = NULL; temp = new Node; temp->next = NULL; p1 = p1->next; p2 = p2->next; for(s=p1;s;s=s->next) { for(r=p2;r;r=r->next) { q = new Node; temp->next = q; q->data = s->data * r->data; q->index = s->index + r->index; q->next = NULL; head = addList(temp,head); } } return head; } //打印多项式 void printList(Node *head) { Node *p = NULL; p = head->next; if(p==NULL) { coutdata>=0) coutdataindex; else coutdataindex; if(p->next!=NULL) coutnext; }while(p != NULL); cout>i; return 0; }
上一篇: jQuery绑定事件的四种方式