本文主要是介绍数据结构之“Ordered List and Sorted List”(七),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
本文主要学习“Sorted List”的应用—— 多项式相加(the addition of two polynomials,点击打开链接)。
一、多项式相加的计算机表示
前面学习“Ordered List”的应用的时候,我们学到用“a sequence of ordered pairs”来表示一个多项式。如下:
然后,用“Ordered List”来表示这个多项式,并编写了一个算法来求该多项式的微分。(点击打开链接)
在计算多项式的微分的算法中,多项式中的每一项在序列中的具体位置并不影响该算法的效率。但是,如果我们考虑两个多项式相加的算法:它需要找出指数(exponent)相同的项,即分组(group),然将指数相同的项的系数(coefficient)相加。如果多项式中的项的位置是任意的,这个分组的过程需要多次遍历“Ordered List”,效率非常低。反之,如果多项式的项是从小到大(按指数)排列的,那么这个分组仅需一次遍历。
二、实现
接口声明:
#pragma once
#include "SortedListAsLinkedList.h"// the oredered pair
class TermB : public Object
{
public:TermB(double, unsigned int);void Put(std::ostream &)const;void Differentiate();double Coefficient() const;unsigned int Exponent() const;friend TermB operator+(const TermB&, const TermB&);protected:int CompareTo(Object const &) const;private:double coefficient;unsigned int exponent;
};class SortedPolynomial : public SortedListAsLinkedList
{
public:SortedPolynomial();~SortedPolynomial();SortedPolynomial(SortedPolynomial&);friend SortedPolynomial operator+(const SortedPolynomial &, const SortedPolynomial &);
};
实现
#include "stdafx.h"
#include "SortedPolynomial.h"TermB::TermB(double _coefficient, unsigned int _exponent): coefficient(_coefficient), exponent(_exponent)
{
}void TermB::Put(std::ostream & s) const
{s << typeid(*this).name() << " {";s << coefficient << "," << exponent;s << " }";
}void TermB::Differentiate()
{if (exponent > 0){coefficient *= exponent;--exponent;}elsecoefficient = 0;
}double TermB::Coefficient() const
{return coefficient;
}unsigned int TermB::Exponent() const
{return exponent;
}TermB operator+(const TermB& arg1, const TermB& arg2)
{if (arg1.exponent != arg2.exponent)throw std::domain_error("unequal exponent");return TermB(arg1.coefficient + arg2.coefficient, arg1.exponent);
}int TermB::CompareTo(Object const & object) const
{TermB const & term = dynamic_cast<TermB const &> (object);if (exponent == term.exponent)return ::Compare(coefficient, term.coefficient);elsereturn exponent - term.exponent;
}SortedPolynomial::SortedPolynomial()
{
}SortedPolynomial::~SortedPolynomial()
{
}SortedPolynomial::SortedPolynomial(SortedPolynomial& arg)
{Purge();Pos & pos = *new Pos(arg, arg.linkedList.Head());//Iterator & pos = arg.NewIterator();while (!pos.IsDone()){const TermB & term = dynamic_cast<const TermB &> (*pos);Insert(*new TermB(term));++pos;}
}SortedPolynomial operator+(const SortedPolynomial & arg1, const SortedPolynomial & arg2)
{SortedPolynomial result;ListAsLinkedList::Pos & pos1 = *new ListAsLinkedList::Pos(arg1, arg1.linkedList.Head());ListAsLinkedList::Pos & pos2 = *new ListAsLinkedList::Pos(arg2, arg2.linkedList.Head());//Iterator & pos1 = arg1.NewIterator();//Iterator & pos2 = arg2.NewIterator();while (!pos1.IsDone() && !pos2.IsDone()){const TermB & term1 = dynamic_cast<const TermB &> (*pos1);const TermB & term2 = dynamic_cast<const TermB &> (*pos2);if (term1.Exponent() < term2.Exponent()){result.Insert(*new TermB(term1));++pos1;}else if (term1.Exponent() > term2.Exponent()){result.Insert(*new TermB(term2));++pos2;}else{TermB sum = term1 + term2;if (sum.Coefficient())result.Insert(*new TermB(sum));++pos1;++pos2;}}while (!pos1.IsDone()){const TermB & term1 = dynamic_cast<const TermB &> (*pos1);result.Insert(*new TermB(term1));++pos1;}while (!pos2.IsDone()){const TermB & term2 = dynamic_cast<const TermB &> (*pos2);result.Insert(*new TermB(term2));++pos2;}delete& pos1;delete& pos2;return result;
}
三、测试
测试代码
// test for Polynomial Addition{SortedPolynomial polynomial;TermB pArray[] = { TermB(5,0), TermB(32,5), TermB(4,2), TermB(56,3), TermB(16,4), TermB(45,1) };for (unsigned int i = 0; i < 6; ++i)polynomial.Insert(pArray[i]);polynomial.Put(std::cout);cout << endl;SortedPolynomial polynomial2;TermB pArray2[] = { TermB(5, 0), TermB(32, 5), TermB(4, 2), TermB(56, 3), TermB(16, 4), TermB(45, 6) };for (unsigned int i = 0; i < 6; ++i)polynomial2.Insert(pArray2[i]);polynomial2.Put(std::cout);cout << endl;SortedPolynomial polynomial3 = polynomial + polynomial2;polynomial3.Put(std::cout);polynomial.RescindOwnership();polynomial2.RescindOwnership();polynomial3.RescindOwnership();}
四、分析和优化
1,“互补多项式”求和
假设计算两个“互补多项式”(a addition of a polynomial and its arithmetic complement)的和。这两个多项式对应项相加为0,最终结果也为零,即在求和函数中无需执行“Insert”。那么它的总消耗时间就是主循环的次数,即O(n)。
注:先考虑特殊情况,再考虑一般情况,这是一种解决问题的思路。
2,一般多项求和
假设两个项数不相等的多项式,p(x) < q(x)。主循环执行L次,次循环执行M次。“Insert”函数为O(k),则求和函数的效率为:
最坏的情况是O(n*n)。主要是因为,我们在分析时假设“Insert”函数不知道每次的具体插入位置。事实上,在循环中,每次都是“在尾部插入”,参照“Insert”函数的实现:
void SortedListAsLinkedList::Insert(Object & object)
{const Node<Object*>* prevPtr = NULL;const Node<Object*>* ptr = linkedList.Head();while (ptr != NULL && *ptr->Datum() < object){prevPtr = ptr;ptr = ptr->Next();}if (!prevPtr)linkedList.Prepend(&object);elselinkedList.InsertAfter(prevPtr, &object);++count;
}
我们可以进行以下优化,取代Insert。
SortedPolynomial operator+(const SortedPolynomial & arg1, const SortedPolynomial & arg2)
{SortedPolynomial result;ListAsLinkedList::Pos & pos1 = *new ListAsLinkedList::Pos(arg1, arg1.linkedList.Head());ListAsLinkedList::Pos & pos2 = *new ListAsLinkedList::Pos(arg2, arg2.linkedList.Head());//Iterator & pos1 = arg1.NewIterator();//Iterator & pos2 = arg2.NewIterator();while (!pos1.IsDone() && !pos2.IsDone()){const TermB & term1 = dynamic_cast<const TermB &> (*pos1);const TermB & term2 = dynamic_cast<const TermB &> (*pos2);if (term1.Exponent() < term2.Exponent()){result.linkedList.Append(new TermB(term1));// result.Insert(*new TermB(term1));++pos1;}else if (term1.Exponent() > term2.Exponent()){result.linkedList.Append(new TermB(term2)); //result.Insert(*new TermB(term2));++pos2;}else{TermB sum = term1 + term2;if (sum.Coefficient())result.linkedList.Append(new TermB(sum)); // result.Insert(*new TermB(sum));++pos1;++pos2;}}while (!pos1.IsDone()){const TermB & term1 = dynamic_cast<const TermB &> (*pos1);result.linkedList.Append(new TermB(term1)); // result.Insert(*new TermB(term1));++pos1;}while (!pos2.IsDone()){const TermB & term2 = dynamic_cast<const TermB &> (*pos2);result.linkedList.Append(new TermB(term2)); // result.Insert(*new TermB(term2));++pos2;}delete& pos1;delete& pos2;return result;
}
优化后,运行时间降为O(n)。
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