链表实践：多项式的加法

使用链表对多项式进行存储和相加，是链表压缩表示能力的一种体现，主要的难点在于对各个项大小的判断，主要的判断逻辑如下（以p、q、r为例，p为第一个多项式链表的第二个节点，q为第二个多项式链表的第二个节点，r为源第一个多项式内存空间新的头结点）：

1. 当p的指数小于q的指数时：将p链接在r的后面。
2. 当p的指数大于q的指数时：将q链接在r的后面。
3. 当p的指数等于q的指数时：a.相加系数和为0，则删去该节点。b.系数不为0，则相加成功。

优化：老师的代码在打印源多项式内容时末尾会出现多余的 “+” ，影响美观，在打印时进行判断可以解决此类情况。

代码设计如下：

//
// Created by Caka on 2023/4/9.
//
#include <iostream>

using namespace std;

/**
 * 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 && p->next!=NULL) {
        cout << p->coefficient << " * " << "10^" << p->exponent << " + ";
        p = p->next;
    }// Of while
    if(p!=NULL){
        cout << p->coefficient << " * " << "10^" << p->exponent;
    }//Of if
    cout << endl;
}// 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) {
        cout << "NULL" << endl;
    } else {
        cout << "The element of " << paraChar << " is (" << paraPtr->coefficient << " * 10^" << paraPtr->exponent << ")"
             << ", the address is " << paraPtr << endl;
    }// 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.
            cout << "Case 1:" << endl;
            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.
            cout << "Case 2:" << endl;
            r->next = q;
            r = q;
            printNode(r, 'r');
            q = q->next;
            printNode(q, 'q');
        } else {
            cout << "Case 3:" << endl;
            //Change the current node of the first list.
            p->coefficient = p->coefficient + q->coefficient;
            cout << "The coefficient is: " << p->coefficient << "." << endl;
            if (p->coefficient == 0) {
                cout << "Case 3.1:" << endl;
                s = p;
                p = p->next;
                printNode(p, 'p');
                // free(s);
            } else {
                cout << "Case 3.2:" << endl;
                r = p;
                printNode(r, 'r');
                p = p->next;
                printNode(p, 'p');
            }// Of if
            s = q;
            q = q->next;
            free(s);
        }// Of if
        cout << "p = " << p << ", q = " << q << endl;
    } // Of while
    cout << "End of while." << endl;

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

    cout << "Addition ends." << endl;
}// 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);
    cout << "The result is: ";
    printList(tempList1);
    cout << endl;
}// 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);
    cout << "The result is: ";
    printList(tempList1);
    cout << endl;
}// Of additionTest2

/**
 * The entrance.
 */
int main() {
    cout << "---PolynomialAddition Test begins---" << endl << endl;
    additionTest1();
    additionTest2();
    cout << "---PolynomialAddition Test ends---";
    return 0;
}// Of main

运行结果如下：

---PolynomialAddition Test begins---

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 address is 0xee1670
The element of q is (8 * 10^1), the address is 0xee1710
The element of r is (0 * 10^0), the address is 0xee1650
Case 1:
The element of r is (7 * 10^0), the address is 0xee1670
The element of p is (3 * 10^1), the address is 0xee1690
p = 0xee1690, q = 0xee1710
Case 3:
The coefficient is: 11.
Case 3.2:
The element of r is (11 * 10^1), the address is 0xee1690
The element of p is (9 * 10^8), the address is 0xee16b0
p = 0xee16b0, q = 0xee1730
Case 2:
The element of r is (22 * 10^7), the address is 0xee1730
The element of q is (-9 * 10^8), the address is 0xee1750
p = 0xee16b0, q = 0xee1750
Case 3:
The coefficient is: 0.
Case 3.1:
The element of p is (5 * 10^17), the address is 0xee16d0
p = 0xee16d0, q = 0
End of while.
Addition ends.
The result is: 7 * 10^0 + 11 * 10^1 + 22 * 10^7 + 5 * 10^17

7 * 10^0 + 3 * 10^1 + 9 * 10^8 + 5 * 10^17
8 * 10^1 + 22 * 10^7 + -9 * 10^10
The element of p is (7 * 10^0), the address is 0xee1710
The element of q is (8 * 10^1), the address is 0xee17d0
The element of r is (0 * 10^0), the address is 0xee16f0
Case 1:
The element of r is (7 * 10^0), the address is 0xee1710
The element of p is (3 * 10^1), the address is 0xee1750
p = 0xee1750, q = 0xee17d0
Case 3:
The coefficient is: 11.
Case 3.2:
The element of r is (11 * 10^1), the address is 0xee1750
The element of p is (9 * 10^8), the address is 0xee1770
p = 0xee1770, q = 0xee17f0
Case 2:
The element of r is (22 * 10^7), the address is 0xee17f0
The element of q is (-9 * 10^10), the address is 0xee18a0
p = 0xee1770, q = 0xee18a0
Case 1:
The element of r is (9 * 10^8), the address is 0xee1770
The element of p is (5 * 10^17), the address is 0xee1790
p = 0xee1790, q = 0xee18a0
Case 2:
The element of r is (-9 * 10^10), the address is 0xee18a0
NULL
p = 0xee1790, q = 0
End of while.
Addition ends.
The result is: 7 * 10^0 + 11 * 10^1 + 22 * 10^7 + 9 * 10^8 + -9 * 10^10 + 5 * 10^17

---PolynomialAddition Test ends---