Abstract : The addition of polynomials is a fundamental application of linked lists and also helps to understand compressed representations.
1. Code (version 2022)
Code first, then nonsense.
#include <stdio.h>
#include <malloc.h>
/**
* 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. Running results
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. Code Description
- I spent a lot of time swiping while writing, so keep all the swipe statements. If you don't like them, you can delete them.
- It is difficult to analyze several situations. Especially after the addition, the coefficient is 0. If you cross this obstacle, you will be advanced!
- After the addition, there is only one linked list left, and other useless spaces are released. If you don't like this method, you can apply for a new space for addition, and the code will be simpler.