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.
1 22 333 44 555 6
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...
}
}
答案 0 :(得分：6)
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;
}
答案 1 :(得分：1)
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();
}
}
答案 2 :(得分：1)
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|
答案 3 :(得分：0)
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;
}
}
}
答案 4 :(得分：0)
这是Kotlin解决方案！
(?#答案 5 :(得分：0)
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]