我必须将多项式表示为arrayslist。如何从类似于此的txt文件中获取输入
P1;5;3;-4;1;8;0
P2;6;5;-2;2;7;1;-4;0
并将其转换为看起来像这样的多项式
P1(X) = 5X^3 –4X +8
P2(X) = 6X^5 -2X^2 +7X -4.
我怎样才能解决这两个多项式之间的加法和减法问题?例如P1 + P2
?
这就是我所拥有的:
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
public class PolyProcessor {
static int polyNum = 0;
public static void main(String[] args) throws FileNotFoundException{
PolyCalc c = new PolyCalc();
File polyfile = new File("polyinput.txt");
Scanner read = new Scanner(polyfile);
while (read.hasNextLine()){
String j = read.nextLine();
c.add(j);
}
}
}
class PolyCalc{
static int polyCount = 0;
static ArrayList polynomials = new ArrayList();
static void add(String j){
polynomials.add(j);
polyCount++;}
static Object get(int i){
return polynomials.get(i);}
}
答案 0 :(得分:3)
多项式加法如何工作?
答案: - 通过添加相同功率的系数
SO P1 = 5X ^ 3 - 4X + 8
且P2 = 6X ^ 5 -2X ^ 2 + 7X ^ 1 + -4
变为
P1 = 0X ^ 5 + 5X ^ 3 + 0X ^ 2 - 4X ^ 1 + 8X ^ 0
P2 = 6X ^ 5 + 0X ^ 3 -2X ^ 2 + 7X ^ 1 - 4X ^ 0
<强> ____________________________________ 强>
SUM = 6X ^ 5 + 5X ^ 3 -2X ^ 2 + 3X ^ 1 + 4X ^ 0
<强> ____________________________________ 强>
您可以将权力存储为Key,将系数作为值存储在Map中。然后迭代地图并在其值中添加系数
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
class SumOfPolynomials {
/**
* @param args
* @throws FileNotFoundException
*/
public static void main(String[] args) throws FileNotFoundException {
List<Map<Integer, Integer>> listOfPolynomials = new ArrayList<Map<Integer, Integer>>();
File polyfile = new File("polyinput.txt");
Scanner read = new Scanner(polyfile);
while (read.hasNextLine()){
String LINE = read.nextLine();
String[] lineSpillted =LINE.split(";");
Map<Integer, Integer> poynomial = new HashMap<Integer, Integer>();
for(int i =1;i<lineSpillted.length-1;i=i+2){ //i starts from ignores P1,P2 etc
poynomial.put(Integer.parseInt(lineSpillted[i+1]), Integer.parseInt(lineSpillted[i]));
}
listOfPolynomials.add(poynomial);
}
read.close();
Map<Integer, Integer> result = polynomialSum(listOfPolynomials.get(0), listOfPolynomials.get(1));
if(listOfPolynomials.size()>2){
for(int i=2;i<listOfPolynomials.size()-1;i++){
result = polynomialSum(result,listOfPolynomials.get(i));
}
}
// print out the SUM as VALUEX^KEY
System.out.println();
int c = 0;
for (Map.Entry<Integer, Integer> entry : result.entrySet()) {
System.out.print(entry.getValue() + "X^" + entry.getKey());
c++;
if (c != result.size()) {
System.out.print("+");
}
}
}
public static Map<Integer, Integer> polynomialSum(Map<Integer, Integer> arg1,
Map<Integer, Integer> arg2) {
Map<Integer, Integer> SUM = new HashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> entry : arg1.entrySet()) {
Integer power = entry.getKey();
Integer coeff1 = entry.getValue();
Integer coefficient;
if (arg2.containsKey(power)) {
coefficient = arg2.get(power) + coeff1;
} else {
coefficient = coeff1;
}
SUM.put(power, coefficient);
}
for (Map.Entry<Integer, Integer> entry : arg2.entrySet()) {
if (SUM.containsKey(entry.getKey())) {
continue;
} else {
SUM.put(entry.getKey(), entry.getValue());
}
}
return SUM;
}
}
已编辑多个多项式。多个多项式在列表中添加,然后通过迭代列表计算总和
<强>输出: - 强>
答案 1 :(得分:1)
以下是关于如何实现多项式的想法:
基于polynomial的定义:
在数学中,多项式是由变量(或不确定)和系数组成的表达式,它只涉及加法,减法,乘法和非负整数指数的运算。
因此,您可以将问题缩减为术语:
class Term {
//making it immutable
final double power;
final double coefficient;
final String variable;
//constructor
public Term(double power, double coefficient, String variable) {
//assign variables and such
this.power = power;
//...
}
//getters for your class
}
现在,创建一个Polynomial
类作为List
个术语,并定义添加和删除术语所需的方法:
class Polynomial {
final String variable;
List<Term> terms;
public Polynomial(String variable) {
//this will allow you to accept only "X" or "Y" or terms with this variable only
this.variable = variable;
terms = new ArrayList<Terms>();
}
public void add(Term term) {
/*
implement this...
*/
}
}
通过这个基本模型,您可以提出更多想法来增强设计。例如,Term
可以实施Comparable<Term>
以支持术语之间的比较,类似于Polynomial
和其他元素。
答案 2 :(得分:0)
我能想到的最简单的解决方案是将系数存储在数组中,并让数组的索引对应x
项上的幂。所以数组:
{2, 4, -1, 1}
会转换为:
x^3 - x^2 + 4x + 2
然后添加和减去只是在两个数组之间添加相应的索引,并将结果存储在新数组中。您还必须跟踪多项式的最高幂项,以便了解如何使用代表它的数组。因此,订单n
的多项式将具有大小为n + 1
的数组来表示它。
答案 3 :(得分:0)
很抱歉变量名称没有接近数学标准,也没有经过测试,但这应该会让你有点想法。
import java.util.ArrayList;
public class Poly {
private String[] numbers;
private ArrayList<Variable> func;
public Poly(String poly, double valueOfX) {
numbers = poly.split(";");
func = new ArrayList<>();
for (int i = 1; i < numbers.length - 1; i+=2) {
double exp = (numbers[i+1] == "0") ? 1 : Double.parseDouble(numbers[i++]);
double x = (numbers[i+1] == "0") ? 1 : valueOfX;
func.add(new Variable(Double.parseDouble(numbers[i]), exp, x));
}
}
public ArrayList<Variable> getFunc() {
return func;
}
}
public class Variable {
private double value;
private double exponent;
private double x;
public Variable(double value, double exponent, double x) {
this.value = value;
this.exponent = exponent;
this.x = x;
}
public double getValue() {
return value;
}
public double getExponent() {
return exponent;
}
public double getX() {
return x;
}
}
由此,您可以获得所需的变量,并通过获取arraylist的索引来计算值,并且可以起到一定的作用。