链表的运用:多项式加法
通过链表来实现两个多项式的加法。
1.创建节点类型
用链表储存多项式则链表的一个节点就代表多项式的某一项。所以一个节点应该包含多项式的系数、多项式的指数以及指向下个节点的指针。
2.打印多项式
传入一个指向多项式链表的指针,遍历该链表,依次打印链表内容,此时采用科学计数法的形式。
3.尾插
申请一个新节点,将要求的系数和指数赋值给这个新节点,并将新节点的指针域置空,然后遍历整个链表,将新节点链接到最后一个节点的next指针上。
4.多项式相加
从两个待相加的多项式表头开始,比较两个节点的指数大小,将结果存储到第一个多项式的链表中。若节点指数相同,则将两个节点系数相加,并将结果存放到第一个节点的系数中;若节点指数不同,将指数小的节点链接到结果链表中,改变指针指向以更新链表遍历状态。
#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;
}// Of initLinkList
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
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 printNode
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;
//Step2: 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
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;
printNode(r, 'r');
free(paraList2);
while ((p != NULL) && (q != NULL))
{
if (p->exponent < q->exponent)
{
//Link the current node of the first list
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)
{
//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 1.
*/
void additionTest1()
{
// 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);
printf("The result is: ");
printList(tempList1);
printf("\r\n");
}// Of additionTest1
/*
*Unit test 2.
*/
void additionTest2()
{
// 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, 10);
printList(tempList2);
// Step 3. Add them to the first.
add(tempList1, tempList2);
printf("The result is: ");
printList(tempList1);
printf("\r\n");
}// Of additionTest2
/*
* The entrance.
*/
int main()
{
additionTest1();
additionTest2();
printf("Finish.\r\n");
return 0;
}// Of main