class多项式（链表实现）

直接上代码是不是不好？

如果是大佬就直接跳过这个阶段。

链表实现的多项式链表

以每一个节点代表多项式的一个项，常数项可以理解为次数为0的项

再用指针将项与项之间联系起来。可以说是数据结构联系必做的东西！

代码390行，慢慢啃吧，有什么不理解，就在评论区问，我还是蛮活跃的。

#include <iostream>
#include <cmath>
#include <sstream>
#include <stack>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;
#define EPS 1e-6struct Node{int exponent;double ratio;Node *next;Node (double r = 0,int e = 0,Node*n = NULL):exponent(e),ratio(r),next(n){};bool operator > (const Node& n) {return exponent > n.exponent; }bool operator < (const Node& n) {return exponent < n.exponent; }bool operator == (const Node& n) {return exponent == n.exponent;}bool operator >= (const Node& n) {return exponent >= n.exponent;}bool operator <= (const Node& n) {return exponent <= n.exponent;}
}; class polynome{Node *first;//判断终止的条件就是在这个链表走到了终点，NULLint size;//记录节点数，用于判断定位删除和添加是否合理
public:polynome();polynome(const polynome & pl);~polynome();polynome& operator =(const polynome & pl);void insert(double r,int e,int pos);void remove(int pos);void clear();int length()const;void sort();//将原数据排序 friend istream& operator >> (istream & cin, polynome & p);friend ostream& operator << (ostream & cout, const polynome & p);//输入输出 //calculate polynome operator + ( polynome & p );polynome operator - ( polynome & p );polynome operator * ( polynome & p );//辅助*polynome help_for_multiply (double r,int e);double getNum(double x);polynome dericative();
}; polynome::polynome():first(NULL),size(0){};
polynome::~polynome() {for(Node *p; p = first; delete p)//思路，如果头指针不是空，就把头指针清掉 first = first -> next;
}
void polynome::insert(double r,int e,int pos) {if(pos > size|| pos < 0)return;if(pos == 0){Node* n = new Node(r,e,first);first = n;++size;return ;} Node *current = first;Node *prev = NULL;int p = 0;for (int i = 0; i < pos;++i){prev = current;current = current->next;}//经过移动，使得将新节点放到prev和current之间 Node *n = new Node(r,e,current);prev->next = n;++size;
}void polynome::remove(int pos) {if (pos < 0|| pos >= size)return;if (pos == 0){Node *p = first;first = first->next;--size;delete p;return;}Node* current = first;Node* prev = NULL;for (int i=0;i < pos;++i){prev = current;current = current->next;}//经过移动，使得要删除的点就是current prev -> next = current ->next;delete current; --size;
}int polynome::length()const {return size;
}void polynome::clear() {for(Node *p; p = first; delete p)//思路，如果头指针不是空，就把头指针清掉 first = first -> next; size = 0;
}polynome::polynome(const polynome & pl) {//copy constructorif(pl.length() <= 0){first = 0;size = 0;return;}*this = pl;
}polynome& polynome::operator = (const polynome & pl) {if (this == &pl)//如果两者在地址上等价 return *this;if(pl.length() <= 0){clear();return *this;}first = new Node(pl.first -> ratio, pl.first -> exponent);Node *current = first;Node *current_pl = pl.first -> next;while (current_pl != NULL){current -> next = new Node(current_pl -> ratio,current_pl -> exponent);current = current -> next;current_pl = current_pl -> next;}size = pl.size;return *this;
}istream& operator >> (istream & cin, polynome & p) {int n,e;double r;p.clear();cin >> n;for (int i = 0; i < n; ++i) {cin >> r>>e;p.insert(r,e,0);}p.sort();return cin;
}ostream& operator << (ostream & cout, const polynome & p) {if (p.length() <= 0)return cout;Node *current = p.first;bool first_time = true;for (int i = 0; i < p.length() ; ++i) {if (current -> ratio == 0) {if (current -> exponent == 0 && p.length() == 1) {cout << current -> ratio;first_time = false;} else {cout << "the exponent =  "<<current->exponent<<endl;cout << "the ratio =  "<<current->ratio<<endl; } current = current -> next;continue;}if (current -> exponent == 0) {if (first_time || current -> ratio < 0) {cout <<current -> ratio;first_time =  false;} else{cout <<"+"<< current -> ratio; } } else if (current -> exponent == 1) {if (first_time || current -> ratio < 0) {if (current -> ratio != 1)cout << current -> ratio << "x";else cout << "x";first_time =  false;} else {cout << "+";if (current -> ratio != 1)cout << current -> ratio << "x";else cout << "x";}} else {if (first_time || current -> ratio < 0){if (current -> ratio != 1)cout << current -> ratio << "x^"<<current->exponent;else cout << "x^"<<current->exponent;first_time =  false;}else{cout << "+";if (current -> ratio != 1)cout << current -> ratio << "x^"<<current->exponent;else cout << "x^"<<current->exponent;}}current = current -> next;}return cout;
} void polynome::sort() { for (Node *p = first; p ; p = p -> next){for (Node* q = p -> next; q; q = q->next) {if (*p > *q) {int temp_exponent = p->exponent;p->exponent = q->exponent;q->exponent = temp_exponent;double temp_ratio = p->ratio;p->ratio = q->ratio;q->ratio = temp_ratio; }}}//完成排序 stack<int>stack_exponent;Node *s,*prev;//用于删除 for (Node *p = first; p ; ) {if (stack_exponent.empty() || stack_exponent.top() != p->exponent){stack_exponent.push(p->exponent);prev = p;p = p -> next; }else {prev->ratio += p->ratio;s = p;prev->next = p->next;p = p->next;delete s;size --;}} //完成去重//完成去不合理的0 prev = first;if (first != NULL){for (Node *p = first -> next; p ; ) {if ( abs(p -> ratio) < EPS ) {prev -> next = p -> next;s = p;p = p -> next;delete s; --size;}else {prev = p;p = p -> next;}}if (size > 1 && abs(first -> ratio) < EPS ) {s = first;first = first -> next;delete s; --size;}}//去除将多项式0转换成数值0if(size == 1 && first -> ratio == 0) {first->exponent = 0;} else if (size == 0) {first = new Node(0,0,0);size ++;}
}polynome polynome::operator + ( polynome & p ) {polynome new_polynome;sort();p.sort();//整理p和this,就是排序，合并重复，还有就是 去掉可以去掉的0项 Node *it_p = first, *it_q = p.first;while (it_p != NULL && it_q != NULL) {if ( it_p -> exponent == it_q -> exponent ) {new_polynome.insert(it_p ->ratio + it_q->ratio, it_p->exponent,0);it_p = it_p -> next;it_q = it_q -> next;} else if (it_p ->exponent < it_q -> exponent ){new_polynome.insert(it_p -> ratio, it_p -> exponent,0);it_p = it_p -> next; } else if (it_p -> exponent > it_q -> exponent) {new_polynome.insert(it_q -> ratio, it_q -> exponent,0);it_q = it_q -> next;}}while (it_p != NULL) {new_polynome.insert(it_p -> ratio, it_p -> exponent,0);it_p = it_p -> next; }while (it_q != NULL) {new_polynome.insert(it_q -> ratio, it_q -> exponent,0);it_q = it_q -> next;}new_polynome.sort();return new_polynome;
}polynome polynome::operator - ( polynome & p ) {polynome new_polynome;sort();p.sort();//整理p和this,就是排序，合并重复，还有就是 去掉可以去掉的0项 Node *it_p = first, *it_q = p.first;while (it_p != NULL && it_q != NULL) {if ( it_p -> exponent == it_q -> exponent ) {new_polynome.insert(it_p ->ratio - it_q->ratio, it_p->exponent,0);it_p = it_p -> next;it_q = it_q -> next;} else if (it_p ->exponent < it_q -> exponent ){new_polynome.insert(it_p -> ratio, it_p -> exponent,0);it_p = it_p -> next; } else if (it_p -> exponent > it_q -> exponent) {new_polynome.insert(-it_q -> ratio, it_q -> exponent,0);it_q = it_q -> next;}}while (it_p != NULL) {new_polynome.insert(it_p -> ratio, it_p -> exponent,0);it_p = it_p -> next; }while (it_q != NULL) {new_polynome.insert(-it_q -> ratio, it_q -> exponent,0);it_q = it_q -> next;}new_polynome.sort();return new_polynome;
}polynome polynome::help_for_multiply (double r,int e) {polynome new_polynome;for (Node *p = first; p ; p = p -> next) {new_polynome.insert( p -> ratio * r, p -> exponent + e, 0);}new_polynome.sort();return new_polynome;
}polynome polynome::operator * ( polynome & p ) {sort();Node *it_p = first;polynome new_polynome,temp;for ( ; it_p ; it_p = it_p -> next) {temp = p.help_for_multiply(it_p -> ratio,it_p -> exponent);new_polynome = new_polynome + temp;}return new_polynome;
}polynome polynome::dericative() {polynome new_polynome;for (Node *p =  first; p ; p = p -> next ) {if (p->exponent == 0){continue;}else {new_polynome.insert(p -> exponent * p -> ratio, p -> exponent - 1, 0);}} new_polynome.sort();return new_polynome;
}double polynome::getNum(double x) {Node *p = first;double sum = 0;while ( p ) {sum +=( pow(x,p->exponent) * p->ratio );p = p -> next;}return sum;
}int main(){polynome p;}

class多项式（链表实现）相关推荐

LeetCode 1634. 求两个多项式链表的和
文章目录 1. 题目 2. 解题 1. 题目多项式链表是一种特殊形式的链表,每个节点表示多项式的一项. 每个节点有三个属性: coefficient:该项的系数.项 9x4 的系数是 9 . pow ...
【数据结构】C++单链表实现多项式加法（直接输入多项式）
题目描述: 计算两个多项式的和. 输入: 输入有两行,每行为一个每项按照指数从大到小顺序的多项式,多项式的每一项用mx^n表示,其中系数m为非0整数,指数n是非负整数. 输出: 输出两个多项式相加的结 ...
MFC版链表实现稀疏多项式相加减
链表实现多项式运算(加减)MFC可视化版题目设计一个一元稀疏多项式简单计算器. 基本要求 (1)输入并建立两个多项式: (2)多项式a与b相加,建立和多项式c: (3)多项式a与b相减,建立差多项 ...
多项式加法运算（链表实现）
文章目录创建结点类型打印多项式尾插插入排序多项式相加代码总览 + 结果展示创建结点类型我们用链表存储一个多项式,那么该链表的每一个结点就代表多项式的某一项.所以我们的每一个结点必须包含 ...
c++用模板实现稀疏多项式_用线性表实现一元多项式及相加运算
" 本文主要讨论线性表在多项式计算中的应用,讨论内容涉及到一元n次多项式在计算机中的表示,及多项式相加运算." 01 在数学上,一个一元n次多项式可以按照升幂写成 Pn(x)= p ...
多项式的链式存储方案
1.题目设计并建立链式存储分配系统,来表示和操作多项式.使用带头结点的循环链表.多项式的每一项表示为一个结点,结点使用如下结构: coef expon link 为了有效地删除多项式,使用可用表空间 ...
数据结构上机实践第四周项目7 - 多项式求和
项目6 - 多项式求和用单链表存储一元多项式,并实现两个多项式的加法. 提示: 1.存储多项式的数据结构多项式的通式是pn(x)=anxn+an−1xn−1+...+a1x+a0.n次多项式共 ...
设计一个一元稀疏多项式简单计算器
目录 1.题目 2.需求分析 3.程序设计 4.测试结果 5.源码 1.题目 [问题描述] 设计一个一元稀疏多项式简单计算器. [基本要求] (1)输入并建立两个多项式: (2)多项式a与b相加,建立 ...

class多项式（链表实现）

class多项式（链表实现）相关推荐

最新文章

热门文章