3.1多项式加法-作业

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#数据结构

本文详细阐述了如何用C语言创建链表表示多项式，并通过两个实例展示了链表节点的添加和多项式相加的过程。

一.代码输入

#include <stdio.h>
#include <malloc.h>


typedef struct LinkNode {
	int coefficient;
	int exponent;
	struct LinkNode* next;
} *LinkList, * NodePtr;


LinkList initLinkList() {
	LinkList tempHeader = (LinkList)malloc(sizeof(struct LinkNode));
	tempHeader->coefficient = 0;
	tempHeader->exponent = 0;
	tempHeader->next = NULL;
	return tempHeader;
}

void printList(LinkList paraHeader) {
	NodePtr p = paraHeader->next;
	while (p != NULL) {
		printf("%d * x^%d + ", p->coefficient, p->exponent);
		p = p->next;
	}
	printf("\r\n");
}


void printNode(NodePtr paraPtr, char paraChar) {
	if (paraPtr == NULL) {
		printf("NULL\r\n");
	}
	else {
		printf("The element of %c is (%d * x^%d)\r\n", paraChar, paraPtr->coefficient, paraPtr->exponent);
	}
}

void appendElement(LinkList paraHeader, int paraCoefficient, int paraExponent) {
	NodePtr p, q;

	q = (NodePtr)malloc(sizeof(struct LinkNode));
	q->coefficient = paraCoefficient;
	q->exponent = paraExponent;
	q->next = NULL;

	p = paraHeader;
	while (p->next != NULL) {
		p = p->next;
	}

	p->next = q;
}

void add(NodePtr paraList1, NodePtr paraList2) {
	NodePtr p, q, r, s;

	p = paraList1->next;
	printNode(p, 'p');
	q = paraList2->next;
	printNode(q, 'q');
	r = paraList1;
	printNode(r, 'r');
	free(paraList2);

	while ((p != NULL) && (q != NULL)) {
		if (p->exponent < q->exponent) {
			printf("case 1\r\n");
			r->next = p;
			r = p;
			printNode(r, 'r');
			p = p->next;
			printNode(p, 'p');
		}
		else if ((p->exponent > q->exponent)) {
			printf("case 2\r\n");
			r->next = q;
			r = q;
			printNode(r, 'r');
			q = q->next;
			printNode(q, 'q');
		}
		else {
			printf("case 3\r\n");
			p->coefficient = p->coefficient + q->coefficient;
			printf("The coefficient is: %d.\r\n", p->coefficient);
			if (p->coefficient == 0) {
				printf("case 3.1\r\n");
				s = p;
				p = p->next;
				printNode(p, 'p');
			}
			else {
				printf("case 3.2\r\n");
				r = p;
				printNode(r, 'r');
				p = p->next;
				printNode(p, 'p');
			}
			s = q;
			q = q->next;
			free(s);
		}

		printf("p = %p, q = %p\r\n", p, q);
	}
	printf("End of while.\r\n");

	if (p == NULL) {
		r->next = q;
	}
	else {
		r->next = p;
	}

	printf("Addition ends.\r\n");
}

void additionTest1() {
	LinkList tempList1 = initLinkList();
	appendElement(tempList1, 7, 0);
	appendElement(tempList1, 3, 1);
	appendElement(tempList1, 9, 8);
	appendElement(tempList1, 5, 17);
	printList(tempList1);

	LinkList tempList2 = initLinkList();
	appendElement(tempList2, 8, 1);
	appendElement(tempList2, 22, 7);
	appendElement(tempList2, -9, 8);
	printList(tempList2);

	add(tempList1, tempList2);
	printf("The result is: ");
	printList(tempList1);
	printf("\r\n");
}

void additionTest2() {
	LinkList tempList1 = initLinkList();
	appendElement(tempList1, 7, 0);
	appendElement(tempList1, 3, 1);
	appendElement(tempList1, 9, 8);
	appendElement(tempList1, 5, 17);
	printList(tempList1);

	LinkList tempList2 = initLinkList();
	appendElement(tempList2, 8, 1);
	appendElement(tempList2, 22, 7);
	appendElement(tempList2, -9, 10);
	printList(tempList2);

	add(tempList1, tempList2);
	printf("The result is: ");
	printList(tempList1);
	printf("\r\n");
}

int main() {
	additionTest1();
	additionTest2();
	printf("Finish.\r\n");
	return 0;
}

2.输出结果

7 * x^0 + 3 * x^1 + 9 * x^8 + 5 * x^17 +
8 * x^1 + 22 * x^7 + -9 * x^8 +
The element of p is (7 * x^0)
The element of q is (8 * x^1)
The element of r is (0 * x^0)
case 1
The element of r is (7 * x^0)
The element of p is (3 * x^1)
p = 0000025C4FADDE70, q = 0000025C4FADDDD0
case 3
The coefficient is: 11.
case 3.2
The element of r is (11 * x^1)
The element of p is (9 * x^8)
p = 0000025C4FADDD30, q = 0000025C4FADDEC0
case 2
The element of r is (22 * x^7)
The element of q is (-9 * x^8)
p = 0000025C4FADDD30, q = 0000025C4FADDF10
case 3
The coefficient is: 0.
case 3.1
The element of p is (5 * x^17)
p = 0000025C4FADDD80, q = 0000000000000000
End of while.
Addition ends.
The result is: 7 * x^0 + 11 * x^1 + 22 * x^7 + 5 * x^17 +

7 * x^0 + 3 * x^1 + 9 * x^8 + 5 * x^17 +
8 * x^1 + 22 * x^7 + -9 * x^10 +
The element of p is (7 * x^0)
The element of q is (8 * x^1)
The element of r is (0 * x^0)
case 1
The element of r is (7 * x^0)
The element of p is (3 * x^1)
p = 0000025C4FADDF60, q = 0000025C4FADDB00
case 3
The coefficient is: 11.
case 3.2
The element of r is (11 * x^1)
The element of p is (9 * x^8)
p = 0000025C4FADE000, q = 0000025C4FADDB50
case 2
The element of r is (22 * x^7)
The element of q is (-9 * x^10)
p = 0000025C4FADE000, q = 0000025C4FADDBA0
case 1
The element of r is (9 * x^8)
The element of p is (5 * x^17)
p = 0000025C4FADE050, q = 0000025C4FADDBA0
case 2
The element of r is (-9 * x^10)
NULL
p = 0000025C4FADE050, q = 0000000000000000
End of while.
Addition ends.
The result is: 7 * x^0 + 11 * x^1 + 22 * x^7 + 9 * x^8 + -9 * x^10 + 5 * x^17 +

Finish.