Java 从二维数组转置矩阵
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Transposing a matrix from a 2D array
提问by Austin Fogg
I'm self teaching myself some java and I'm stuck on creating a 2D array that initializes it with random values and then creates the transpose of the array.
我正在自学一些 Java，我一直坚持创建一个二维数组，用随机值初始化它，然后创建数组的转置。
An example output is:
一个示例输出是：
$ 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.
Original matrix
原始矩阵
1 22 333 44 555 6
1 22 333 44 555 6
Transposed matrix
转置矩阵
1 333 555` 22 44 6`
1 333 555` 22 44 6`
^ Should be the final output. Some help with the code would appreciated!
^ 应该是最终输出。对代码的一些帮助将不胜感激！
I would like to code to generate error messages if the number of rows or columns is outside the specified range. And for if to read the matrix elements from the command line and not generate them randomly.
如果行数或列数超出指定范围，我想编写代码以生成错误消息。以及 if 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...
}
}
采纳答案by thebesttony
This is a simple method that return an int[][] of the transposed matrix...
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;
}
Than to print a 2D matrix you can use a method like this:
比打印 2D 矩阵，您可以使用这样的方法：
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;
}
回答by Ashish K Gupta
The answer provided above is not efficient in terms of memory. It is using another array - transposedMatrix apart from array supplied as argument. This will lead to consume double memory. We can do this in-place as follows:
上面提供的答案在内存方面效率不高。除了作为参数提供的数组之外，它使用另一个数组 - transposedMatrix。这会导致消耗双倍内存。我们可以按如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();
}
}
回答by Vineet Abbi
You can use the below class it has most of the methods you want.
您可以使用下面的类，它具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());
}
}
Output:
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|
回答by Piyush Shandilya
For a square matrix, instead of iterating through the entire array, you just iterate through the diagonally half of the 2D array and swap the values with the corresponding indices.
对于方阵，不是遍历整个数组，而是遍历 2D 数组的对角线一半并将值与相应的索引交换。
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;
}
}
}
