我有一个基于行的多维数组：
/** [row][column]. */
public int[][] tiles;
我想将此数组转换为基于列的数组，如下所示：
/** [column][row]. */
public int[][] tiles;
...但我真的不知道从哪里开始
问问题
50636 次
11 回答
17
我看到所有答案都创建了一个新的结果矩阵。这很简单：
matrix[i][j] = matrix[j][i];
但是，如果是方阵，您也可以就地执行此操作。
// Transpose, where m == n
for (int i = 0; i < m; i++) {
for (int j = i + 1; j < n; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
这对于较大的矩阵更好，因为创建一个新的结果矩阵在内存方面是浪费的。如果它不是方形的，您可以创建一个具有NxM尺寸的新的并执行不合适的方法。注意：对于就地，照顾j = i + 1。不是0。
于 2013-10-13T05:31:29.480 回答
10
下面的解决方案实际上返回了一个转置数组，而不仅仅是打印它，并且适用于所有矩形数组，而不仅仅是正方形。
public int[][] transpose(int[][] array) {
// empty or unset array, nothing do to here
if (array == null || array.length == 0)
return array;
int width = array.length;
int height = array[0].length;
int[][] array_new = new int[height][width];
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
array_new[y][x] = array[x][y];
}
}
return array_new;
}
例如，您应该通过以下方式调用它：
int[][] a = new int[][]{{1, 2, 3, 4}, {5, 6, 7, 8}};
for (int i = 0; i < a.length; i++) {
System.out.print("[");
for (int y = 0; y < a[0].length; y++) {
System.out.print(a[i][y] + ",");
}
System.out.print("]\n");
}
a = transpose(a); // call
System.out.println();
for (int i = 0; i < a.length; i++) {
System.out.print("[");
for (int y = 0; y < a[0].length; y++) {
System.out.print(a[i][y] + ",");
}
System.out.print("]\n");
}
这将按预期输出：
[1,2,3,4,]
[5,6,7,8,]
[1,5,]
[2,6,]
[3,7,]
[4,8,]
于 2013-07-13T20:38:30.600 回答
8
试试这个：
@Test
public void transpose() {
final int[][] original = new int[][]{
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}};
for (int i = 0; i < original.length; i++) {
for (int j = 0; j < original[i].length; j++) {
System.out.print(original[i][j] + " ");
}
System.out.print("\n");
}
System.out.print("\n\n matrix transpose:\n");
// transpose
if (original.length > 0) {
for (int i = 0; i < original[0].length; i++) {
for (int j = 0; j < original.length; j++) {
System.out.print(original[j][i] + " ");
}
System.out.print("\n");
}
}
}
输出：
1 2 3 4
5 6 7 8
9 10 11 12
matrix transpose:
1 5 9
2 6 10
3 7 11
4 8 12
于 2011-12-07T21:09:56.883 回答
3
更通用的方式：
/**
* Transposes the given array, swapping rows with columns. The given array might contain arrays as elements that are
* not all of the same length. The returned array will have {@code null} values at those places.
*
* @param <T>
* the type of the array
*
* @param array
* the array
*
* @return the transposed array
*
* @throws NullPointerException
* if the given array is {@code null}
*/
public static <T> T[][] transpose(final T[][] array) {
Objects.requireNonNull(array);
// get y count
final int yCount = Arrays.stream(array).mapToInt(a -> a.length).max().orElse(0);
final int xCount = array.length;
final Class<?> componentType = array.getClass().getComponentType().getComponentType();
@SuppressWarnings("unchecked")
final T[][] newArray = (T[][]) Array.newInstance(componentType, yCount, xCount);
for (int x = 0; x < xCount; x++) {
for (int y = 0; y < yCount; y++) {
if (array[x] == null || y >= array[x].length) break;
newArray[y][x] = array[x][y];
}
}
return newArray;
}
于 2017-01-04T09:40:39.950 回答
1
如果您想对矩阵进行就地转置（在这种情况下row count = col count），您可以在 Java 中进行以下操作
public static void inPlaceTranspose(int [][] matrix){
int rows = matrix.length;
int cols = matrix[0].length;
for(int i=0;i<rows;i++){
for(int j=i+1;j<cols;j++){
matrix[i][j] = matrix[i][j] + matrix[j][i];
matrix[j][i] = matrix[i][j] - matrix[j][i];
matrix[i][j] = matrix[i][j] - matrix[j][i];
}
}
}
于 2014-01-26T15:19:28.713 回答
0
public int[][] tiles, temp;
// Add values to tiles, wherever you end up doing that, then:
System.arraycopy(tiles, 0, temp, 0, tiles.length);
for (int row = 0; row < tiles.length; row++) // Loop over rows
for (int col = 0; col < tiles[row].length; col++) // Loop over columns
tiles[col][row] = temp[row][col]; // Rotate
那应该为你做。
于 2011-12-07T20:59:30.307 回答
0
import java.util.Arrays;
import java.util.Scanner;
public class Demo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
//asking number of rows from user
int rows = askArray("Enter number of rows :", input);
//asking number of columns from user
int columns = askArray("Enter number of columns :", input);
int[][] array = Array(rows, columns, input);
//displaying initial matrix
DisplayArray(array, rows, columns);
System.out.println("Transpose array ");
//calling Transpose array method
int[][] array2 = TransposeArray(array, rows, columns);
for (int i = 0; i < array[0].length; i++) {
System.out.println(Arrays.toString(array2[i]));
}
}
//method to take number of rows and number of columns from the user
public static int askArray(String s, Scanner in) {
System.out.print(s);
int value = in.nextInt();
return value;
}
//feeding elements to the matrix
public static int[][] Array(int x, int y, Scanner input) {
int[][] array = new int[x][y];
for (int j = 0; j < x; j++) {
System.out.print("Enter row number " + (j + 1) + ":");
for (int i = 0; i < y; i++) {
array[j][i] = input.nextInt();
}
}
return array;
}
//Method to display initial matrix
public static void DisplayArray(int[][] arra, int x, int y) {
for (int i = 0; i < x; i++) {
System.out.println(Arrays.toString(arra[i]));
}
}
//Method to transpose matrix
public static int[][] TransposeArray(int[][] arr, int x, int y) {
int[][] Transpose_Array = new int[y][x];
for (int i = 0; i < x; i++) {
for (int j = 0; j < y; j++) {
Transpose_Array[j][i] = arr[i][j];
}
}
return Transpose_Array;
}
}
于 2015-12-29T23:52:30.530 回答
0
public int[][] getTranspose() {
int[][] transpose = new int[row][column];
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
transpose[i][j] = original[j][i];
}
}
return transpose;
}
于 2016-11-24T09:14:16.703 回答
0
使用此函数（如有必要，将 String 替换为 int）。它将矩阵作为字符串数组并返回一个新矩阵，即转置。它也检查空数组的边缘情况。没有指纹。
private String[][] transposeTable(String[][] table) {
// Transpose of empty table is empty table
if (table.length < 1) {
return table;
}
// Table isn't empty
int nRows = table.length;
int nCols = table[0].length;
String[][] transpose = new String[nCols][nRows];
// Do the transpose
for (int i = 0; i < nRows; i++) {
for (int j = 0; j < nCols; j++) {
transpose[j][i] = table[i][j];
}
}
return transpose;
}
于 2019-11-27T19:20:14.353 回答
0
这是我的 50 美分：一种转换多维数组的实用方法和测试（在我的例子中是双精度数）：
/**
* Transponse bidimensional array.
*
* @param original Original table.
* @return Transponsed.
*/
public static double[][] transponse(double[][] original) {
double[][] transponsed = new double
[original[0].length]
[original.length];
for (int i = 0; i < original[0].length; i++) {
for (int j = 0; j < original.length; j++) {
transponsed[i][j] = original[j][i];
}
}
return transponsed;
}
@Test
void aMatrix_OfTwoDimensions_ToBeTransponsed() {
final double[][] original =
new double[][]{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
double[][] transponsed = Analysis.transponse(original);
assertThat(transponsed[1][2], is(equalTo(10)));
}
于 2020-03-09T10:23:49.460 回答
-1
import java.util.*;
public class TestClass {
public static void main(String args[]) throws Exception {
Scanner in = new Scanner(System.in);
int iSize = in.nextInt();
int jSize = in.nextInt();
int arr[][] = new int[iSize][jSize];
int array[][] = new int[jSize][iSize];
for (int i = 0; i < iSize; i++) {
for (int j = 0; j < jSize; j++) {
arr[i][j] = in.nextInt();
}
System.out.println("\n");
}
for (int n = 0; n < arr[0].length; n++) {
for (int m = 0; m < arr.length; m++) {
array[n][m] = arr[m][n];
System.out.print(array[n][m] + " ");
}
System.out.print("\n");
}
}
}
于 2019-09-26T02:26:29.453 回答