0.618法

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一种经典的一维搜索方法

用的集合装，实际应该更简单！

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package cn.sgnxotsmicf.Demo;
import java.text.DecimalFormat;
import java.time.LocalDateTime;
import java.time.temporal.ChronoUnit;
import java.util.ArrayList;
import java.util.Scanner;
public class Demo_Number {
    static	double T = 0.618;
    static {
        System.out.println("======0.618法求解最优化问题======");
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        ArrayList<Double> Data = new ArrayList<>();
        inPut(sc,Data);
    }

    private static void inPut(Scanner sc,ArrayList<Double> Data) {
        System.out.println("请输入允许误差ε:");
        double c = sc.nextDouble();
        System.out.println("给定搜索左区间a:");
        double a = sc.nextInt();
        System.out.println("给定搜索右区间b:");
        double b = sc.nextInt();
        LocalDateTime startTime = LocalDateTime.now();
        Init(Data,a,b);
        Judge(Data,c);
        LocalDateTime endTime = LocalDateTime.now();
        System.out.println("迭代时间大约为:"+ChronoUnit.MICROS.between(startTime, endTime)+"微秒");
    }

    /**
     ！！！注意：函数在此方法下修改
     */
    private static double function(double tempt){
        return tempt*tempt-tempt+2;//0.554
    }

    private static void Init(ArrayList<Double> Data,double a, double b) {
        double L = a+(1-T)*(b-a);
        double U = a+T*(b-a);
        double L1 = function(L);
        double U1 = function(U);
        Data.add(L);//0
        Data.add(U);//1
        Data.add(L1);//2
        Data.add(U1);//3
        Data.add(a);//4
        Data.add(b);//5
    }

    private static void Judge(ArrayList<Double> Data,double c) {
        while(true) {
            if(Math.abs(Data.get(4)-Data.get(5))>c) {
                if(Data.get(2)<Data.get(3)) {
                    Data.set(5, Data.get(1));//b的值
                    Data.set(1, Data.get(0));
                    Data.set(3, Data.get(2));
                    Data.set(0, Data.get(4)+(1-T)*(Data.get(5)-Data.get(4)));
                    double tempt = Data.set(0, Data.get(4)+(1-T)*(Data.get(5)-Data.get(4)));
                    Data.set(2, function(tempt));
                }else {
                    Data.set(4, Data.get(0));//a的值
                    Data.set(0,Data.get(1));
                    Data.set(2,Data.get(3));
                    Data.set(1,Data.get(4)+T*(Data.get(5)-Data.get(4)));
                    double tempt = Data.set(1,Data.get(4)+T*(Data.get(5)-Data.get(4)));
                    Data.set(3,function(tempt));
                }
            }else {
                DecimalFormat sc = new DecimalFormat("0.000");
                System.out.println("最优解x="+sc.format((Data.get(4)+Data.get(5))/2));
                break;
            }
        }
    }
}