Java 3D数组找到最近的点？

Question

正在尝试查找从具有“坐标 [x,y]s”的 12 个元素的 3D 数组生成的最近点：

   public int[] nearSquare (int xPos, int yPos) {
      int len = 40;
      int[][][] loc = new int[12][12][2];
      int[] output = new int[2];
      int boundary = len/2;
      for(int size = 0,i = len/2; size < 12; size ++,i += len) {
          for(int size2 = 0,j = len/2; size2 < 12; size2 ++,j += len) {
              loc[size][size2][0] = j;
              loc[size][size2][1] = i;
          }
      }
      for (int row = 0; row < loc.length; row++) {
          for (int col = 0; col < loc[row].length; col++) {
              if (loc[row][col][0] - boundary < xPos && loc[row][col][0] + boundary > xPos && loc[row][col][1] - boundary < yPos && loc[row][col][1] + boundary > yPos) {
                      output[0] = loc[row][col][0]; 
                      output[1] = loc[row][col][1];
              } else {
                  continue;
              }
          }
      }
      return output;
  }

该方法应该生成中心坐标的 3D 数组，然后将它们与输入值 xPos 和 yPos 进行比较，以找到最近的点并将其作为数组输出。图中所示方法的假定概念：

红点作为输入的xPos和yPos。尝试找到最近的中心点（在 3D 数组中生成）（绿色）然后输出其坐标

问题是，在尝试运行一个 Java @test 时（nearSquare 方法在 class 中称为“shapeTest” ):

public class Test
{
   shapeTest st;
   public Test() {}
   @BeforeEach
   public void setUp()
   {
       st = new shapeTest();
   }
   // Trying to test here...
   @Test
   public void testnearSquare() {
       int[] test1 = new int[2];
       int[] test1val = st.nearSquare(0,0);
       assertEquals(test1[0], test1val[0]);
       assertEquals(test1[1], test1val[1]);
    
       int[] test2 = new int[2];
       test2[1] = 7;
       int[] test2val = st.nearSquare(0,300);
       assertEquals(test2[0], test2val[0]);
       assertEquals(test2[1], test2val[1]);
   }
}

测试将失败说明expected: <7> but was: <0>和org.opentest4j.AssertionFailedError: expected: <7> but was: <0>如果是由于“nearSquare”方法没有正确过滤最近的坐标，输出最近点的正确方法是什么输入坐标？

Answer 1

据我了解（我可能是错的），您遇到的是 2D 问题而不是 3D 问题，因为您进入该方法的点没有 z-coordinate.

Given point (x,y), determine nearest point (x',y') with

x' in {20, 60, 100, 140, ..., 460} or n * 40 + 20 with n in [0,...,11]
and
y' in {20, 60, 100, 140, ..., 460} or m * 40 + 20 with m in [0,...,11]

但是您的测试和方法名称表明您要确定 n 和 m:

nearSquare(0,0) -> (0,0)
nearSquare(0,300) -> (0,7)

所以让我们编写代码：

import java.util.*;
public class NearSquare {
    static final int squareLength = 40;
    static final int numberOfSquares = 12; // in one dimension

    public static int[] nearSquare(int xPos, int yPos) {
        int n, m;

        // determine n
        if (xPos <= 0) {
            n = 0; // min index
        } else if (xPos >= squareLength * numberOfSquares) {
            n = numberOfSquares - 1; // max index
        } else {
            n = xPos / squareLength;
        }

        // determine m
        if (yPos <= 0) {
            m = 0; // min index
        } else if (yPos >= squareLength * numberOfSquares) {
            m = numberOfSquares - 1; // max index
        } else {
            m = (yPos - 20) / squareLength;
        }
        return new int[] {n, m};
    }

    public static void main(String[] args) 
    {
        int square1[] = nearSquare(0, 0);
        System.out.println("Test 1: nearSquare(0,0) -> (0,0): " + (square1[0] == 0 && square1[1] == 0 ? "passed" : "failed"));
        System.out.println("Result: " + Arrays.toString(square1));
        System.out.println("Nearest center point: ("
            + (square1[0]*squareLength+squareLength/2)
            + ", "
            + (square1[1]*squareLength+squareLength/2)
            + ")"
        );

        int square2[] = nearSquare(0, 300);
        System.out.println("Test 2: nearSquare(0,0) -> (0,7): " + (square2[0] == 0 && square2[1] == 7 ? "passed" : "failed"));
        System.out.println("Result: " + Arrays.toString(square2));
        System.out.println("Nearest center point: ("
            + (square2[0]*squareLength+squareLength/2)
            + ", "
            + (square2[1]*squareLength+squareLength/2)
            + ")"
        );
    }
}

并展示它给 table 带来的好处：

$ java NearSquare.java
Test 1: nearSquare(0,0) -> (0,0): passed
Result: [0, 0]
Nearest center point: (20, 20)
Test 2: nearSquare(0,0) -> (0,7): passed
Result: [0, 7]
Nearest center point: (20, 300)
$

Java 3D数组找到最近的点？

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