Java在坐标系中找出距离最近的点

Dear 丶 2023-09-30 13:51 23阅读 0赞

在有限点中找出距离最近的点，要求用二维数组且为浮点型（Double)编写，编写一个distance方法计算距离在主方法中来比较

输入：输入点的个数和坐标点（不需要括号和逗号）

8
    
     -1 3 -1 -1 1 1 2 0.5 2 -1 3 3 4 2 4 -0.5

输出：输出距离最近的点的坐标

Enter the number of points: 
    Enter   points:
    
    The closest two points are (1.0,1.0) and (2.0,0.5)

import java.util.Scanner;
    
    public class FindNearestPoints {
        
        public static void main(String[] args) {
        
            Scanner input = new Scanner(System.in);
            System.out.print("Enter the number of points: ");
            int numberOfPoints = input.nextInt();
    
            //Create an array to store points
            double[][] points = new double[numberOfPoints][2];
            System.out.print("Enter " + numberOfPoints + "points: ");
            for (int i = 0; i < points.length; i++) {
        
                points[i][0] = input.nextDouble();
                points[i][1] = input.nextDouble();
            }
    
            //p1 and p2 are the indices in the points array
            int p1 = 0, p2 = 1;
            double shortestDistance = distance(points[p1][0],points[p1][1],points[p2][0],points[p2][1]);
    
            //Compute distance for every two points
            for (int i = 0; i < points.length; i++) {
        
                for (int j = i + 1; j < points.length; j++) {
        
                    double distance = distance(points[i][0],points[i][1],points[j][0],points[j][1]);
                    if (shortestDistance > distance){
        
                        p1 = i;
                        p2 = j;
                        shortestDistance = distance;
                    }
                }
            }
    
            //Display result
            System.out.println("The closest two points are " + "(" + points[p1][0] + ","
             + points[p1][1] + ") and (" + points[p2][0] + "," + points[p2][1] + ")");
        }
    
        public static double distance(double x1,double y1,double x2,double y2){
        
            return Math.sqrt((x2 - x1)*(x2 - x1) + (y2-y1)*(y2-y1));
        }
    }