PAT 1002 A+B for Polynomials

问题：A、B两个多项式求和，结果按要求输出，考虑到小数点后一位。原问题链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805526272000000

解决方案：

这是一个可以用链表（List）作为数据结构，使用归并排序解决的问题。首先定义链表中的节点（Node），Node中成员变量有exponent和coefficient，以及指向下一个节点的next指针；然后利用定义好的Node构造List，List中的成员变量有Node的数目,以及head，tail指针，除了成员变量之外，还需要定义成员函数，List的成员函数有insert，search，delete，printf等等，这一题只需定义insert和prinft即可；最后根据输入生成两个链表，使用归并排序即可，注意要考虑到两个元素相加和为0的情况。

C++代码：

#include<iostream>
#include<iomanip>　　　　//题目中要求考到的小数点后一位，故调用这个头文件

using namespace std;

class Node{        //定义Node，注意成员变量尽量用private，成员函数用public
private:
    int exponent;
    float coefficient;
    Node* next;

public:
    Node(int exponent = 0, float coefficient = 0.0, Node* next = NULL) {
        this->exponent = exponent;
        this->coefficient = coefficient;
        this->next = next;
    }
    int getExpon() {
        return this->exponent;
    }
    float getCoeff() {
        return this->coefficient;
    }
    void setExpon(int exponent) {
        this->exponent = exponent;
    }
    void setCoeff(float coefficient) {
        this->coefficient = coefficient;
    }
    Node* getNext() {
        return this->next;
    }
    void setNext(Node* A) {
        this->next = A;
    }
    ~Node() {
        delete this->next;
    }
};

class List {
private:
    int number;
    Node* head, * tail;

public:
    List() {
        this->number = 0;
        this->head = this->tail = new Node(-1, 0);    //这里有个链表的trick，前面加一个没有数据的head，便于后续insert等操作
    }
    int getNumber() {
        return this->number;
    }
    void addNumber() {
        this->number++;
    }
    void minusNumber() {
        this->number--;
    }
    Node* getHead() {
        return this->head;
    }
    Node* getTail() {
        return this->tail;
    }
    void setHead(Node* A) {
        this->head = A;
    }
    void setTail(Node* A) {
        this->tail = A;
    }
    void insert(int exponent, float coefficient);        //声明insert函数
    void printf();                                        //声明printf函数
    ~List() {
        delete this->head;
        delete this->tail;
    }
};

void List::insert(int exponent, float coefficient) {
    Node* A = new Node(exponent, coefficient);
    Node* B = this->head;
    for (; B->getNext() != NULL && A->getExpon() < B->getNext()->getExpon(); B = B->getNext());
    if (B->getNext() == NULL) {        //直接插在最后
        this->getTail()->setNext(A);
        this->setTail(A);
        this->addNumber();
    }
    else if(A->getExpon() == B->getNext()->getExpon()){        //这里有两种情况，一种是和不为0，一种是和为0
        if (B->getNext()->getCoeff() + A->getCoeff() != 0)    //和不为0
            B->getNext()->setCoeff(B->getNext()->getCoeff() + A->getCoeff());
        else {        //和为0，消掉链表中的一个Node，注意这个Node正好是在最后的情况，注意出理tail指针
            Node* C = B->getNext();
            B->setNext(B->getNext()->getNext());
            if (B->getNext() == NULL)        //处理tail指针
                this->setTail(B);
            this->minusNumber();
            delete(A);
            delete(C);
        }
    }
    else if(A->getExpon() > B->getNext()->getExpon()) {        //插在中间，这就体现了链表前面有一个空head的好处了
        A->setNext(B->getNext());
        B->setNext(A);
        this->addNumber();
    }
}

void List::printf() {
    cout << this->getNumber();
    for (Node* A = this->getHead()->getNext(); A != NULL; A = A->getNext()) {
        cout << " " << A->getExpon();
        cout << fixed << setprecision(1) << " " << A->getCoeff();　　　　//考虑到小数点后一位，使用"iomanip"头文件
    }

}

int main() {
    int number, exponent;
    float coefficient;
    List* A, * B, * C;
    Node* NA, * NB;
    A = new List();
    B = new List();
    C = new List();
    cin >> number;
    for (int i = 0; i < number; i++) {
        cin >> exponent >> coefficient;
        A->insert(exponent, coefficient);
    }
    cin >> number;
    for (int i = 0; i < number; i++) {
        cin >> exponent >> coefficient;
        B->insert(exponent, coefficient);
    }
    NA = A->getHead()->getNext();
    NB = B->getHead()->getNext();
    while (NA != NULL && NB != NULL) {        //归并排序，注意和为0的情况
        if (NA->getExpon() < NB->getExpon()) {
            C->insert(NB->getExpon(), NB->getCoeff());
            NB = NB->getNext();
        }
        else if (NA->getExpon() == NB->getExpon()) {
            if (NA->getCoeff() + NB->getCoeff() != 0)    //和不为0
                C->insert(NA->getExpon(), NA->getCoeff() + NB->getCoeff());
            NA = NA->getNext();
            NB = NB->getNext();
        }
        else {
            C->insert(NA->getExpon(), NA->getCoeff());
            NA = NA->getNext();
        }
    }
    for (; NA != NULL; NA = NA->getNext()) C->insert(NA->getExpon(), NA->getCoeff());
    for (; NB != NULL; NB = NB->getNext()) C->insert(NB->getExpon(), NB->getCoeff());
    C->printf();
    return 0;
}