Java矩阵 - 转置&总和

时间:2015-06-19 15:58:51

标签: java arrays eclipse matrix transpose

我正在开发一个关于eclipse的Matrix程序,目前我仍然坚持使用2种方法,其中我认为这是最简单的方法。我正在研究的第一种方法是从@Test案例中获取2个不同2D数组的总和,并在新的2D数组中返回总和。我已经有了一个2D数组实例变量。我被卡住的原因是因为方法中的参数。该参数不提供除类(Matrix)和变量(其他)之外的任何变量。所以我想知道如何开始这个方法,最重要的是返回sum数组。

我坚持的另一种方法是转置方法,你必须翻转给定2D数组的行和列。我知道我必须创建一个临时2D数组,以便将内容存储回原始的2D数组,但由于某种原因,它没有通过测试用例。如果有人可以帮我解决这两种方法,我们将不胜感激。

import java.util.Arrays;

public class Matrix {

    private int[][] array;
    private int[][] array2;
    private int theRow;
    private int theCol;

    public Matrix(int[][] arrayOfArrays) {
        // TODO Auto-generated constructor stub
        array = new int[arrayOfArrays.length][arrayOfArrays[0].length];

        for (int r = 0; r < arrayOfArrays.length; r++) {
            for (int c = 0; c < arrayOfArrays[r].length; c++) {
                array[r][c] = arrayOfArrays[r][c];
            }
        }

    }

    public int get(int row, int column) {

        return array[row][column];
    }


    public int getNumberOfRows() {

    int nRows = array.length;

    return nRows;
    }


    public int getNumberOfColumns() {

    int nCols = array[0].length;

    return nCols;


    }
    public String toString() {
        String res = "";
        for (int r = 0; r < array.length; r++) {
            for (int c = 0; c < array[r].length; c++) 
                res = res + array[r][c];
        }

        return res;
    }
    public Matrix sum(Matrix other) {


        return sum;
    }
    public void scalarMultiply(int scalar) {

        for (int r = 0; r < array.length; r++)  {
            for (int c = 0; c < array[0].length; c++) {
                array[r][c] = array[r][c] * scalar;
            }
        }
    }

    public void transpose() {

        int m = array.length;
        int n = array[0].length;

        int[][] transpose = new int [n][m];

        int temp;
        for (int r = 0; r < m; r++)  {
            for (int c = 0; c < n; c++)  {
                transpose[c][r] = array[r][c];
                array[r][c] = array[c][r];
                array[c][r] = transpose[c][r];                      
            }
        }
    }

    //The test cases for sum method and transpose method

    @Test
    public void testSum() {
        int[][] a1 = { { 1, 2, 3 }, 
                       { 5, 6, 7 } };
        Matrix a = new Matrix(a1);

        int[][] a2 = { { -2, -2, -2 }, 
                       { 4, 4, 4 } };
        Matrix b = new Matrix(a2);

        Matrix c = a.sum(b);

        assertEquals(-1, c.get(0, 0));
        assertEquals(0, c.get(0, 1));
        assertEquals(1, c.get(0, 2));
        assertEquals(9, c.get(1, 0));
        assertEquals(10, c.get(1, 1));
        assertEquals(11, c.get(1, 2));
    }

    @Test
    public void testTranspose() {

        int[][] a1 = { { 1, 3, 5 },
                       { 2, 4, 6 } };
        Matrix a = new Matrix(a1);
        a.transpose();

        assertEquals(1, a.get(0, 0));
        assertEquals(2, a.get(0, 1));
        assertEquals(3, a.get(1, 0));
        assertEquals(4, a.get(1, 1));
        assertEquals(5, a.get(2, 0));
        assertEquals(6, a.get(2, 1));
    }

1 个答案:

答案 0 :(得分:2)

您需要更改尺寸,例如,2x3 - &gt; 3×2。

import java.util.Arrays;

public class Matrix {

    private int[][] array;
    private int[][] array2;// remove this
    private int theRow;// remove this
    private int theCol;// remove this


    public void transpose() {

        int m = array.length;
        int n = array[0].length;

        int[][] transpose = new int [n][m];

        for (int r = 0; r < m; r++)  {
            for (int c = 0; c < n; c++)  {
                transpose[c][r] = array[r][c];                     
            }
        }
        array = transpose;
    }
}