算法－第四版－练习1.2.3解答

编写一个Interval2D的用例，从命令行接受参数N、min和max。生成N个随机的2D间隔，其宽度和高均匀地分布在单位正方形中的min和max之间。用StdDraw画出它们并打印出相交的间隔对的数量以及有包含关系的间隔对数量。

/**
 * Description : 
 * Author      : [e-mail]
 * Date        : Sep 26, 2016 3:26:18 PM
 * Copyright (c) 2013-2016, http://furzoom.com All Rights Reserved.
 */
package com.furzoom.lab.algs.ch102;

import edu.princeton.cs.algs4.Interval1D;
import edu.princeton.cs.algs4.Interval2D;
import edu.princeton.cs.algs4.StdRandom;

/**
 * ClassName    : E10203 <br>
 * Function     : TODO ADD FUNCTION. <br>
 * date         : Sep 26, 2016 3:26:18 PM <br>
 * 
 * @version 
 */
public class E10203
{
    public static double min(double d1, double d2)
    {
        return d1 < d2 ? d1 : d2;
    }
    
    public static double max(double d1, double d2)
    {
        return d1 > d2 ? d1 : d2;
    }
    
    public static void main(String[] args)
    {
        int n = Integer.parseInt(args[0]);
        double min = Double.parseDouble(args[1]);
        double max = Double.parseDouble(args[2]);
        
        Interval2D[] intervals = new Interval2D[n];
        for (int i = 0; i < n; i++)
        {
            double d1 = StdRandom.uniform(min, max);
            double d2 = StdRandom.uniform(min, max);
            double d3 = StdRandom.uniform(min, max);
            double d4 = StdRandom.uniform(min, max);
   
            intervals[i] = new Interval2D(new Interval1D(min(d1, d2), max(d1, d2)),
                    new Interval1D(min(d3, d4), max(d3, d4)));
        }
        // draw all Interval2D
        for (int i = 0; i < n; i++)
        {
            intervals[i].draw();
        }
        // 
        int countIntersects = 0;
        int countContains = 0;
        for (int i = 0; i < n - 1; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                if (intervals[i].intersects(intervals[j]))
                {
                    countIntersects++;
                }
//                if (intervals[i].contains(intervals[j]))
//                {
//                    
//                }
            }
        }
        System.out.println(countIntersects);
        
    }
}

结果如下：

相交的数量为5.

算法－第四版－1.2 数据抽象－习题索引汇总

算法－第四版习题索引汇总