我有一个基于行的多维数组:
/** [row][column]. */
public int[][] tiles;
我想将这个数组转换为基于列的数组,就像下面这样:/** [column][row]. */
public int[][] tiles;
...但我真的不知道从哪里开始
我有一个基于行的多维数组:
/** [row][column]. */
public int[][] tiles;
我想将这个数组转换为基于列的数组,就像下面这样:/** [column][row]. */
public int[][] tiles;
...但我真的不知道从哪里开始
我看到所有的答案都创建了一个新的结果矩阵。这很简单:
matrix[i][j] = matrix[j][i];
然而,如果是方阵的情况下,你也可以在原地进行此操作。
// 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;
}
}
对于较大的矩阵而言,这种方法更为优越,因为创建一个新的结果矩阵会浪费内存。如果它不是方阵,您可以创建一个 NxM
维度的新矩阵并使用 out-of-place 方法。注意:对于 in-place 方法,请确保 j = i + 1
而不是 0
。
以下解决方案实际上返回转置数组而不仅仅是打印它,并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;
}
你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,]
试一下这个:
@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
/**
* 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;
}
如果你想进行矩阵的原地转置(在这种情况下,行数=列数
),可以按以下方式使用Java:
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];
}
}
}
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;
}
}
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;
}
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
这应该对你有用。
temp = tiles
会使tiles
和temp
成为同一个数组的引用,而你已经完全搞砸了它。:-P - ruakhtmp
应该是 temp
。然后,您从未实例化 temp
(我假设 tiles
来自某些外部来源)。但即使修复了这些问题,它也会将 [[1, 2], [3, 4]]
转换为 [[1, 2], [2, 4]]
。如果矩阵不是正方形而是矩形(即 N x M 而不是 N x N),它将抛出 ArrayIndexOutOfBoundsException。 - yshavit这是我的建议:提供一个实用方法和测试来转置多维数组(在我的情况下是双精度):
/**
* 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)));
}
这是我大约一年前想出来的,当时我刚开始学习编程,所以在学习Java两周后就想出了解决方案。
我们得到一个值列表:
int[] jest = new int[] {1, 2, 3, 4, 5, 6};
然后我们生成空的双重列表,获取单个列表的长度,并将它们存储在双重列表中。
int[][] storeValues = null;
int[][] valueLength = new int[jest.length][jest.length];
int[][] submit = null;
int[][] submitted = null;
for(int i=0; i<jest.length;i++) {
for(int j = 0; j < jest.length;j++) {
storeValues = new int[][] {jest};
valueLength[j][i] = storeValues[0][i];
submit = Arrays.copyOfRange(valueLength, 0, j+1);
submitted = Arrays.copyOfRange(submit, j, j+1);
}
完整的工作代码:
import java.util.Arrays;
public class transMatrix {
public static void main(String[] args) {
int[] jest = new int[] {1, 2, 3, 4, 5, 6};
int[][] storeValues = null;
int[][] valueLength = new int[jest.length][jest.length];
int[][] submit = null;
int[][] submitted = null;
for(int i=0; i<jest.length;i++) {
for(int j = 0; j < jest.length;j++) {
storeValues = new int[][] {jest};
valueLength[j][i] = storeValues[0][i];
submit = Arrays.copyOfRange(valueLength, 0, j+1);
submitted = Arrays.copyOfRange(submit, j, j+1);
}
}System.out.println(Arrays.deepToString(submitted));
}}
输出:
[[1, 2, 3, 4, 5, 6]]