数据结构–链表实现多项式求和&&求积
- 首先,数据的组织形式:使用数组来存储结构体数据相对比较简单。但是为了熟悉链表的使用,这里使用链表。
- 多项式的表示:输入的时候先输入系数,然后输入指数。输出同样是这样。
难点:
- 求多项式乘积不像求和,因为后插入数据的指数可能比前一个数据要大,所以需要进行判断。
#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
#include <stdlib.h>
typedef struct PolyNode* Polynomial;
struct PolyNode
{
int coef;
int expon;
Polynomial link;
};
void attach(int c, int e, Polynomial* pRear)
{
Polynomial P = (Polynomial)malloc(sizeof(struct PolyNode));
P->coef = c;
P->expon = e;
P->link = NULL;
(*pRear)->link = P;
*pRear = P;
}
Polynomial readPoly()
{
int num1 = 0, c = 0, e = 0;
printf("请输入多项式的系数:");
scanf("%d", &num1);
Polynomial Head = (Polynomial)malloc(sizeof(struct PolyNode));
Polynomial Rear = Head;
Head->link = NULL;
while (num1--)
{
scanf("%d %d", &c, &e);
attach(c, e, &Rear);
}
Polynomial T = Head;
Head = Head->link;
free(T);
return Head;
}
int compare(int a, int b)
{
if (a > b) return 1;
else if (a == b) return 0;
else return -1;
}
Polynomial polyAdd(Polynomial P1, Polynomial P2)
{
Polynomial Head, Rear, T;
Head = (Polynomial)malloc(sizeof(struct PolyNode));
Head->link = NULL;
Rear = Head;
int sum = 0;
while (P1 && P2)
{
switch (compare(P1->expon, P2->expon))
{
case 1:
attach(P1->coef, P1->expon, &Rear);
P1 = P1->link;
break;
case 0:
sum = P1->coef + P2->coef;
if (sum) attach(sum, P1->expon, &Rear);
P1 = P1->link; P2 = P2->link;
break;
case -1:
attach(P2->coef, P2->expon, &Rear);
P2 = P2->link;
break;
}
}
for (; P1; P1 = P1->link) attach(P1->coef, P1->expon, &Rear);
for (; P2; P2 = P2->link) attach(P2->coef, P2->expon, &Rear);
T = Head;
Head = Head->link;
free(T);
return Head;
}
void printPoly(Polynomial P)
{
int flag = 0;
while (P)
{
if (flag)
{
printf(" ");
}
else
{
flag = 1;
}
printf("%d %d", P->coef, P->expon);
P = P->link;
}
}
Polynomial polyMulti(Polynomial P1, Polynomial P2)
{
Polynomial T1, T2, Head, Rear, P;
Head = (Polynomial)malloc(sizeof(struct PolyNode));
Head->link = NULL;
Rear = Head;
T1 = P1;
while (T1)
{
T2 = P2; Rear = Head;
while (T2)
{
int c = T1->coef * T2->coef;
int e = T1->expon + T2->expon;
while (Rear->link && Rear->link->expon > e)
{
Rear = Rear->link;
}
if (Rear->link && Rear->link->expon == e)
{
if (c + Rear->link->coef) Rear->link->coef = c + Rear->link->coef;
else
{
Polynomial tmp = Rear->link;
Rear->link = tmp->link;
free(tmp);
}
}
else
{
P = (Polynomial)malloc(sizeof(struct PolyNode));
P->coef = c;
P->expon = e;
P->link = Rear->link;
Rear->link = P;
Rear = Rear->link;
}
T2 = T2->link;
}
T1 = T1->link;
}
Polynomial ttt = Head;
Head = ttt->link;
free(ttt);
return Head;
}
int main()
{
Polynomial P1 = readPoly();
Polynomial P2 = readPoly();
Polynomial P3 = polyAdd(P1, P2);
printPoly(P3);
Polynomial P4 = polyMulti(P1, P2);
printf("\n");
printPoly(P4);
return 0;
}
结果如下
如求多项式: 3x4+5x2+6x1+2 和 5x20+7x4+3x 的乘积 和是:5x20+10x4+5x2+9x+2 乘积为:15x24+25x22+30x21+…
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