数据结构 C 代码 2.5: 多项式的加法

摘要: 多项式的加法是链表的基本应用, 也有助于理解压缩表示.

1. 代码 (2022 版)

先上代码, 再说废话.

#include 
#include 

/**
 * Linked list of integers. The key is data. The key is sorted in non-descending order.
 */
typedef struct LinkNode{
	int coefficient;
	int exponent;
	struct LinkNode *next;
} *LinkList, *NodePtr;

/**
 * Initialize the list with a header.
 * @return The pointer to the header.
 */
LinkList initLinkList(){
	LinkList tempHeader = (LinkList)malloc(sizeof(struct LinkNode));
	tempHeader->coefficient = 0;
	tempHeader->exponent = 0;
	tempHeader->next = NULL;
	return tempHeader;
}// Of initLinkList

/**
 * Print the list.
 * @param paraHeader The header of the list.
 */
void printList(LinkList paraHeader){
	NodePtr p = paraHeader->next;
	while (p != NULL) {
		printf("%d * 10^%d + ", p->coefficient, p->exponent);
		p = p->next;
	}// Of while
	printf("\r\n");
}// Of printList

/**
 * Print one node for testing.
 * @param paraPtr The pointer to the node.
 * @param paraChar The name of the node.
 */
void printNode(NodePtr paraPtr, char paraChar){
	if (paraPtr == NULL) {
		printf("NULL\r\n");
	} else {
		printf("The element of %c is (%d * 10^%d)\r\n", paraChar, paraPtr->coefficient, paraPtr->exponent);
	}// Of while
}// Of printNode

/**
 * Add an element to the tail.
 * @param paraCoefficient The coefficient of the new element.
 * @param paraExponent The exponent of the new element.
 */
void appendElement(LinkList paraHeader, int paraCoefficient, int paraExponent){
	NodePtr p, q;

	// Step 1. Construct a new node.
	q = (NodePtr)malloc(sizeof(struct LinkNode));
	q->coefficient = paraCoefficient;
	q->exponent = paraExponent;
	q->next = NULL;

	// Step 2. Search to the tail.
	p = paraHeader;
	while (p->next != NULL) {
		p = p->next;
	}// Of while

	// Step 3. Now add/link.
	p->next = q;
}// Of appendElement

/**
 * Polynomial addition.
 * @param paraList1 The first list.
 * @param paraList2 The second list.
 */
void add(NodePtr paraList1, NodePtr paraList2){
	NodePtr p, q, r, s;

	// Step 1. Search to the position.
	p = paraList1->next;
	printNode(p, 'p');
	q = paraList2->next;
	printNode(q, 'q');
	r = paraList1; // Previous pointer for inserting.
	printNode(r, 'r');
	free(paraList2); // The second list is destroyed. 
	
	while ((p != NULL) && (q != NULL)) {
		if (p->exponent < q->exponent) {
			//Link the current node of the first list.
			printf("case 1\r\n");
			r = p;
			printNode(r, 'r');
			p = p->next;
			printNode(p, 'p');
		} else if ((p->exponent > q->exponent)) {
			//Link the current node of the second list.
			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");
			//Change the current node of the first list.
			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');
				// free(s);
			} else {
				printf("case 3.2\r\n");
				r = p;
				printNode(r, 'r');
				p = p->next;
				printNode(p, 'p');
			}// Of if
			s = q;
			q = q->next;
			//printf("q is pointing to (%d, %d)\r\n", q->coefficient, q->exponent);
			free(s);
		}// Of if

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

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

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

/**
 * Unit test.
 */
void additionTest(){
	// Step 1. Initialize the first polynomial.
	LinkList tempList1 = initLinkList();
	appendElement(tempList1, 7, 0);
	appendElement(tempList1, 3, 1);
	appendElement(tempList1, 9, 8);
	appendElement(tempList1, 5, 17);
	printList(tempList1);

	// Step 2. Initialize the second polynomial.
	LinkList tempList2 = initLinkList();
	appendElement(tempList2, 8, 1);
	appendElement(tempList2, 22, 7);
	appendElement(tempList2, -9, 8);
	printList(tempList2);

	// Step 3. Add them to the first.
	add(tempList1, tempList2);
	printList(tempList1);
}// Of additionTest

/**
 * The entrance.
 */
void main(){
	additionTest();
	printf("Finish.\r\n");
}// Of main

2. 运行结果

7 * 10^0 + 3 * 10^1 + 9 * 10^8 + 5 * 10^17 +
8 * 10^1 + 22 * 10^7 + -9 * 10^8 +
The element of p is (7 * 10^0)
The element of q is (8 * 10^1)
The element of r is (0 * 10^0)
case 1
The element of r is (7 * 10^0)
The element of p is (3 * 10^1)
p = 5652160, q = 5652384
case 3
The coefficient is: 11.
case 3.2
The element of r is (11 * 10^1)
The element of p is (9 * 10^8)
p = 5652216, q = 5641768
case 2
The element of r is (22 * 10^7)
The element of q is (-9 * 10^8)
p = 5652216, q = 5641824
case 3
The coefficient is: 0.
case 3.1
The element of p is (5 * 10^17)
p = 5652272, q = 0
End of while.
Addition ends.
7 * 10^0 + 11 * 10^1 + 22 * 10^7 + 5 * 10^17 +
Finish.
Press any key to continue

3. 代码说明

1. 我在写的时候都用了不少时间调拭, 所以将所有的调拭语句保留. 如果不喜欢, 可以将它们删除.
2. 对几种情况的分析有一定难度. 特别是相加后系数为 0. 跨越这个障碍就会进阶!
3. 相加后仅剩一个链表, 其它无用空间都被释放. 如果不喜欢这种方式, 可以申请新空间进行相加, 这样的代码其实会更简单.