英文:
Sorting 2D array of integers by column
问题
我需要在Java中构建一个方法,其中输入是一个整数的二维数组,结果是一个整数的二维数组,其中每个元素都引用列中的一个元素位置。让我用一个示例来解释。将一个5x5的二维数组作为方法的输入,如下所示:
int[][] array = new int[][]{
{124, 188, 24, 254, 339},
{0, 7, 77, 145, 159},
{206, 340, 280, 523, 433},
{310, 265, 151, 411, 398},
{24, 104, 0, 183, 198}};
现在,我需要根据以下方式构建一个新的整数的二维数组(以下将其称为 newArray
):
-
列 0 中的最小值是 0,与行 1 相关联(因此,我需要在
newArray[0][0]
中分配 1)。 -
接下来,列 0 中的最小值是 24,与行 4 相关联(因此,我需要在
newArray[1][0]
中分配 4)。 -
然后,列 0 中的最小值是 124,与行 0 相关联(因此,我需要在
newArray[2][0]
中分配 0)。 -
对于每一列依此类推...
该方法的最终输出应该类似于以下二维数组。
英文:
I need to build a method in Java where the input is a 2D array of integers and get as a result a 2D array of integers where each element makes reference to a position of an element in a column. Let me explain that with an example. Consider as an input for the method a 2D arrays of 5x5 as follow:
int[][] array = new int[][]{
{124, 188, 24, 254, 339},
{0, 7, 77, 145, 159},
{206, 340, 280, 523, 433},
{310, 265, 151, 411, 398},
{24, 104, 0, 183, 198}};
Now I need to build a new 2D array of integer (I will call newArray
in the following) according to:
-
The minimum value of column 0 in the array is 0 and is associated with row 1 (then, I need to assign a 1 in
newArray[0][0]
). -
Next, the minimum value of column 0 in the array is 24 and is associated with row 4 (then, I need to assign a 4 in
newArray[1][0]
). -
Then, the minimum value of column 0 in the array is 124 and is associated with row 0 (then, I need to assign a 0 in
newArray[2][0]
). -
And so on for each column...
The final output of the method must be something like the following 2d array.
Any help would be highly appreciated.
答案1
得分: 1
如果我理解正确:
输入:
{{124,188,24,254,339},
{0,7,77,145,159},
{206,340,280,523,433},
{310,265,151,411,398},
{24,104,0,183,198}}
输出:
{{1,1,4,1,1}
{4,4,0,4,4}
{0,0,1,0,0}
{2,3,3,3,3}
{3,2,2,2,2}}
以下是代码:
public static int[][] createArray(int[][] a) {
int[][] nA = new int[a.length][a[0].length];
int[] col = new int[a.length];
int minIndex = -1;
for (int i = 0; i < a.length; i++) {
// 首先取出列
for (int j = 0; j < a[0].length; j++) {
col[j] = a[j][i];
}
// 遍历列
for (int k = 0; k < a[0].length; k++) {
int min = Integer.MAX_VALUE;
// 遍历列中剩余的数字
for (int j = 0; j < col.length; j++) {
// 找到剩余数字中的最小值
if (min > col[j]) {
min = col[j];
minIndex = j;
}
}
// 从数组中删除该数字
col[minIndex] = Integer.MAX_VALUE;
// 在最终数组中设置这个数字
nA[k][i] = minIndex;
}
}
return nA;
}
可能有更简单的方法,但这个方法是有效的!
英文:
If I understood correctly :
IN :
{{124, 188, 24, 254, 339},
{0, 7, 77, 145, 159},
{206, 340, 280, 523, 433},
{310, 265, 151, 411, 398},
{24, 104, 0, 183, 198}}
OUT :
{{1, 1, 4, 1, 1}
{4, 4, 0, 4, 4}
{0, 0, 1, 0, 0}
{2, 3, 3, 3, 3}
{3, 2, 2, 2, 2}
Here's the code :
public static int[][] createArray(int[][] a) {
int[][] nA = new int[a.length][a[0].length];
int[] col = new int[a.length];
int minIndex = -1;
for (int i = 0; i < a.length; i++) {
// First get the col out
for (int j = 0; j < a[0].length; j++) {
col[j] = a[j][i];
}
// Loop through the col
for (int k = 0; k < a[0].length; k++) {
int min = Integer.MAX_VALUE;
// Loop through the remaining numbers of the col
for (int j = 0; j < col.length; j++) {
// Find the remaining lowest number
if (min > col[j]) {
min = col[j];
minIndex = j;
}
}
// Delete the number from the array
col[minIndex] = Integer.MAX_VALUE;
// Set this number in the final array
nA[k][i] = minIndex;
}
}
return nA;
}
There might be an easier way, but it works !
答案2
得分: 0
按列排序矩阵元素的索引,实际上是通过转置矩阵的行来对元素的索引进行排序:
int m = 5;
int n = 6;
int[][] arr1 = new int[][]{
{124, 188, 24, 254, 339, 3},
{0, 7, 77, 145, 159, 1},
{206, 340, 280, 523, 433, 5},
{310, 265, 151, 411, 398, 4},
{24, 104, 0, 183, 198, 2}};
int[][] arr2 = IntStream
// 遍历数组的行索引
.range(0, n)
.mapToObj(i -> IntStream
// 遍历数组的列索引
.range(0, m)
.boxed()
// 根据数组中的值对列的索引进行排序
.sorted(Comparator.comparingInt(j -> arr1[j][i]))
.mapToInt(Integer::intValue)
// 排序后的索引列即为新数组的一行
.toArray())
// 返回排序后的索引数组
.toArray(int[][]::new);
// 转置索引数组
int[][] arr3 = new int[m][n];
IntStream.range(0, m).forEach(i ->
IntStream.range(0, n).forEach(j ->
arr3[i][j] = arr2[j][i]));
// 输出
Arrays.stream(arr3).map(Arrays::toString).forEach(System.out::println);
输出结果:
[1, 1, 4, 1, 1, 1]
[4, 4, 0, 4, 4, 4]
[0, 0, 1, 0, 0, 0]
[2, 3, 3, 3, 3, 3]
[3, 2, 2, 2, 2, 2]
英文:
Sorting the indexes of matrix elements by columns is sorting the indexes of elements by the rows of a transposed matrix:
int m = 5;
int n = 6;
int[][] arr1 = new int[][]{
{124, 188, 24, 254, 339, 3},
{0, 7, 77, 145, 159, 1},
{206, 340, 280, 523, 433, 5},
{310, 265, 151, 411, 398, 4},
{24, 104, 0, 183, 198, 2}};
int[][] arr2 = IntStream
// iterate over the indices
// of the rows of the array
.range(0, n)
.mapToObj(i -> IntStream
// iterate over the
// indices of the columns
.range(0, m)
.boxed()
// sort indices of the elements of the
// columns by its values in the array
.sorted(Comparator.comparingInt(j -> arr1[j][i]))
.mapToInt(Integer::intValue)
// sorted column of indices is
// a row in the new array
.toArray())
// return sorted array of indices
.toArray(int[][]::new);
// transpose the array of indices
int[][] arr3 = new int[m][n];
IntStream.range(0, m).forEach(i ->
IntStream.range(0, n).forEach(j ->
arr3[i][j] = arr2[j][i]));
// output
Arrays.stream(arr3).map(Arrays::toString).forEach(System.out::println);
Output:
[1, 1, 4, 1, 1, 1]
[4, 4, 0, 4, 4, 4]
[0, 0, 1, 0, 0, 0]
[2, 3, 3, 3, 3, 3]
[3, 2, 2, 2, 2, 2]
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