Question

我自学了一些java，我不得不创建一个用随机值初始化它的2D数组，然后创建数组的转置。

示例输出是：

$ java Test1 22 333 44 555 6  
Enter the number of rows (1-10): 0  
ERROR: number not in specified range (1-10) !  
and so on until you enter the correct number of rows and columns.

原始矩阵

  1  22  
333  44  
555   6

转置矩阵

  1  333  555`  
 22   44    6`

^应该是最终输出。一些帮助代码将不胜感激！

如果行数或列数超出指定范围，我想编码生成错误消息。并且if是从命令行读取矩阵元素而不是随机生成它们。

import java.util.Scanner;

public class Test1 {
    /** Main method */
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter the number of rows (1-10): ");
        int rows = input.nextInt();
        System.out.print("Enter the number of columns (1-10): ");
        int cols = input.nextInt();

        // Create originalMatrix as rectangular two dimensional array
        int[][] originalMatrix = new int[rows][cols];

        // Assign random values to originalMatrix
        for (int row = 0; row < originalMatrix.length; row++)
            for (int col = 0; col < originalMatrix[row].length; col++) {
                originalMatrix[row][col] = (int) (Math.random() * 1000);
            }

        // Print original matrix
        System.out.println("\nOriginal matrix:");
        printMatrix(originalMatrix);

        // Transpose matrix
        int[][] resultMatrix = transposeMatrix(originalMatrix);


        // Print transposed matrix
        System.out.println("\nTransposed matrix:");
        printMatrix(resultMatrix);
    }

    /** The method for printing the contents of a matrix */
    public static void printMatrix(int[][] matrix) {
        for (int row = 0; row < matrix.length; row++) {
            for (int col = 0; col < matrix[row].length; col++) {
                System.out.print(matrix[row][col] + "  ");
            }
            System.out.println();
        } 
    }

    /** The method for transposing a matrix */
    public static int[][] transposeMatrix(int[][] matrix) {
        // Code goes here...
    }
}

Answer 1

这是一个返回转置矩阵的int [] []的简单方法......

public static int[][] transposeMatrix(int[][] matrix){
    int m = matrix.length;
    int n = matrix[0].length;

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

    for(int x = 0; x < n; x++) {
        for(int y = 0; y < m; y++) {
            transposedMatrix[x][y] = matrix[y][x];
        }
    }

    return transposedMatrix;
}

要打印2D矩阵，您可以使用以下方法：

public static String matrixToString(int[][] a){
    int m = a.length;
    int n = a[0].length;

    String tmp = "";
    for(int y = 0; y<m; y++){
        for(int x = 0; x<n; x++){
            tmp = tmp + a[y][x] + " ";
        }
        tmp = tmp + "\n";
    }

    return tmp;
}

Answer 2

上面提供的答案在内存方面效率不高。它正在使用另一个数组 - transposedMatrix除了作为参数提供的数组。这将导致消耗双倍内存。我们可以按照以下方式就地执行此操作：

public void transposeMatrix(int[][] a)
{
        int temp;
        for(int i=0 ; i<(a.length/2 + 1); i++)
        {
            for(int j=i ; j<(a[0].length) ; j++)
            {
                temp = a[i][j];
                a[i][j] = a[j][i];
                a[j][i] = temp;
            }
        }

        displayMatrix(a);
    }

public void displayMatrix(int[][] a){
        for(int i=0 ; i<a.length ; i++)
        {
            for(int j=0 ; j<a[0].length ; j++)
            {
                System.out.print(a[i][j] + " ");
            }

            System.out.println();
        }
    }

Answer 3

您可以使用以下类，它拥有您想要的大部分方法。

/**
 * Class representing square matrix of given size.
 * It has methods to rotate by 90, 180 and 270
 * And also to transpose and flip on either axis.
 * 
 * I have used both space efficient methods in transpose and flip
 * And simple but more space usage for rotation. 
 * 
 * This is using builder pattern hence, you can keep on applying
 * methods say rotate90().rotate90() to get 180 turn.
 * 
 */
public class Matrix {

    private int[][] matrix;
    final int size;

    public Matrix(final int size) {
        this.size = size;
        matrix = new int[size][size];

        for (int i=0;i<size;i++)
            for (int j=0;j<size;j++)
                matrix[i][j] = i*size + j;
    }

    public Matrix rotate90() {
        int[][] temp = new int[size][size];

        for (int i=0;i<size;i++)
            for (int j=0;j<size;j++)
                temp[i][j] = matrix[size-1-j][i];

        matrix = temp;
        return this;
    }
    public Matrix rotate180() {
        int[][] temp = new int[size][size];

        for (int i=0;i<size;i++)
            for (int j=0;j<size;j++)
                temp[i][j] = matrix[size-1-i][size-1-j];

        matrix = temp;
        return this;
    }
    public Matrix rotate270() {
        int[][] temp = new int[size][size];

        for (int i=0;i<size;i++)
            for (int j=0;j<size;j++)
                temp[i][j] = matrix[j][size-1-i];

        matrix = temp;
        return this;
    }
    public Matrix transpose() {
        for (int i=0; i<size-1; i++) {
            for (int j=i+1; j<size; j++) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = tmp;
            }
        }
        return this;
    }
    public Matrix flipVertical() {
        for (int i=0; i<size; i++) {
            for (int j=0; j<size/2; j++) {
                int tmp = matrix[i][size-1-j];
                matrix[i][size-1-j] = matrix[i][j];
                matrix[i][j] = tmp;
            }
        }
        return this;
    }
    public Matrix flipHorizontal() {
        for (int i=0; i<size/2; i++) {
            for (int j=0; j<size; j++) {
                int tmp = matrix[size-1-i][j];
                matrix[size-1-i][j] = matrix[i][j];
                matrix[i][j] = tmp;
            }
        }
        return this;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        for (int i=0;i<size;i++) {
            for (int j=0;j<size;j++) {
                sb.append("|");
                sb.append(matrix[i][j]);
                if (size > 3) {
                    sb.append("\t");
                }
            }
            sb.append("|\n");
        }

        return sb.toString();
    }

    public static void main(String... args) {
        Matrix m = new Matrix(3);
        System.out.println(m);

        //transpose and flipHorizontal is 270 turn (-90)
        System.out.println(m.transpose());
        System.out.println(m.flipHorizontal());

        //rotate 90 further to bring it back to original position
        System.out.println(m.rotate90());

        //transpose and flip Vertical is 90 degree turn
        System.out.println(m.transpose().flipVertical());
    }
}

输出：

|0|1|2|
|3|4|5|
|6|7|8|

|0|3|6|
|1|4|7|
|2|5|8|

|2|5|8|
|1|4|7|
|0|3|6|

|0|1|2|
|3|4|5|
|6|7|8|

|6|3|0|
|7|4|1|
|8|5|2|

Answer 4

对于方形矩阵，您只需迭代2D数组的对角线一半，然后将值与相应的索引交换，而不是遍历整个数组。

public void transposeMatrix(int[][] a) {
        for(int i=0 ; i<n; i++) { 
            for(int j=0 ; j<i ; j++) {
                int temp = a[i][j];
                a[i][j] = a[j][i];
                a[j][i] = temp;
            }
        }
}

Answer 5

这是Kotlin解决方案！

(?#

Answer 6

n x m 矩阵的另一种 Kotlin 解决方案

fun transpose(matrix: Array<Array<Double>>): Array<Array<Double>> {
    return matrix[0].mapIndexed { col, _ ->
        matrix.mapIndexed { row, _ ->
            matrix[row][col]
        }.toTypedArray()
    }.toTypedArray()
}

fun printMatrix(matrix: Array<Array<Double>>) {
    println(matrix.joinToString("\n") { it.contentToString() })
}

fun main() {
    val matrix = arrayOf(
        arrayOf( 1.0,  2.0,  3.0),
        arrayOf( 4.0,  5.0,  6.0),
        arrayOf( 7.0,  8.0,  9.0),
        arrayOf( 1.0,  2.0,  3.0),
    )
    println("matrix:")
    printMatrix(matrix)
    println("transpose(matrix):")
    printMatrix(transpose(matrix))
}

输出

matrix:
[1.0, 2.0, 3.0]
[4.0, 5.0, 6.0]
[7.0, 8.0, 9.0]
[1.0, 2.0, 3.0]
transpose(matrix):
[1.0, 4.0, 7.0, 1.0]
[2.0, 5.0, 8.0, 2.0]
[3.0, 6.0, 9.0, 3.0]

在Kotlin Playground

中尝试一下

从2D阵列转置矩阵

原始矩阵

转置矩阵

6 个答案: