英文:
Insert column in desired location of 2D array
问题
我想选择数组中的一列,比如第2列。我希望将这一列插入到二维数组的特定位置,比如第4列。
例如:
1 3 5 5 2
2 4 6 2 1
3 6 9 1 1
期望的输出将是:
1 5 5 3 2
2 6 2 4 1
3 9 1 6 1
我知道我可以使用以下代码循环,直到逐个交换每一列,直到列在所需位置为止:
for (int[] array1 : array) {
int temp = array1[col1];
array1[col1] = array1[col2];
array1[col2] = temp;
}
然而,如果我使用宽度为30列的大矩阵,这将非常低效。是否有一种方法可以在二维数组中的任何位置插入列,而不必迭代每次交换直到它在正确的位置?
英文:
I'm looking to choose a column of my array, lets say column 2. I want this column to be inserted at a specific location in the 2D array, lets say column 4.
For example:
1 3 5 5 2
2 4 6 2 1
3 6 9 1 1
The desired output would be:
1 5 5 3 2
2 6 2 4 1
3 9 1 6 1
I know I could loop the following code until I 1 by 1 swap every column until the column is in the desired location.
for (int[] array1 : array) {
int temp = array1[col1];
array1[col1] = array1[col1];
array1[col2] = temp;
}
However, if I'm using large matrices such as 30 columns wide, this would be incredibly inefficient. Is there a way to insert the column anywhere in the 2D array without iterating through each swap until it is in the right spot?
答案1
得分: 2
可能通过使用流进行并行处理来提高性能。
应该使用IntStream
的行索引来单独处理每一行。
// 基本的位移方法
// from, to - 从1开始的索引
public static void shiftArray(int[] arr, int from, int to) {
int tmp = arr[from - 1];
for (int i = from; i < to; i++) {
arr[i - 1] = arr[i];
}
arr[to - 1] = tmp;
}
public static void main(String[] args) {
int[][] arr2d = {
{1, 3, 5, 5, 2},
{2, 4, 6, 2, 1},
{3, 6, 9, 1, 1}
};
int fromColumn = 2;
int toColumn = 4;
IntStream.range(0, arr2d.length)
.parallel()
.forEach(i -> shiftArray(arr2d[i], fromColumn, toColumn)); // 并行处理每一行
// 输出位移后的二维数组
Arrays.stream(arr2d)
.map(Arrays::toString)
.forEach(System.out::println);
}
输出:
[1, 5, 5, 3, 2]
[2, 6, 2, 4, 1]
[3, 9, 1, 6, 1]
英文:
Possibly, a performance may be improved by using parallel processing with streams.
IntStream
of row indexes should be used to handle each row separately.
// basic shift method
// from, to - indexes starting from 1
public static void shiftArray(int[] arr, int from, int to) {
int tmp = arr[from - 1];
for (int i = from; i < to; i++) {
arr[i - 1] = arr[i];
}
arr[to - 1] = tmp;
}
public static void main(String[] args) {
int[][] arr2d = {
{1, 3, 5, 5, 2},
{2, 4, 6, 2, 1},
{3, 6, 9, 1, 1}
};
int fromColumn = 2;
int toColumn = 4;
IntStream.range(0, arr2d.length)
.parallel()
.forEach(i -> shiftArray(arr2d[i], fromColumn, toColumn)); // shift each row in parallel
// print the 2D array after shift
Arrays.stream(arr2d)
.map(Arrays::toString)
.forEach(System.out::println);
}
Output:
[1, 5, 5, 3, 2]
[2, 6, 2, 4, 1]
[3, 9, 1, 6, 1]
答案2
得分: 1
尝试这个。
public static void moveColumn(int[][] matrix, int from, int to) {
--from; --to; // 如果列号从零开始,请删除此行。
int srcPos = from < to ? from + 1 : to;
int destPos = from < to ? from : to + 1;
int length = Math.abs(from - to);
for (int[] array : matrix) {
int temp = array[from];
System.arraycopy(array, srcPos, array, destPos, length);
array[to] = temp;
}
}
和
int[][] matrix = {
{1, 3, 5, 5, 2},
{2, 4, 6, 2, 1},
{3, 6, 9, 1, 1}
};
moveColumn(matrix, 2, 4);
for (int[] row : matrix)
System.out.println(Arrays.toString(row));
输出
[1, 5, 5, 3, 2]
[2, 6, 2, 4, 1]
[3, 9, 1, 6, 1]
英文:
Try this.
public static void moveColumn(int[][] matrix, int from, int to) {
--from; --to; // If column number begins with zero, remove this line.
int srcPos = from < to ? from + 1 : to;
int destPos = from < to ? from : to + 1;
int length = Math.abs(from - to);
for (int[] array : matrix) {
int temp = array[from];
System.arraycopy(array, srcPos, array, destPos, length);
array[to] = temp;
}
}
and
int[][] matrix = {
{1, 3, 5, 5, 2},
{2, 4, 6, 2, 1},
{3, 6, 9, 1, 1}
};
moveColumn(matrix, 2, 4);
for (int[] row : matrix)
System.out.println(Arrays.toString(row));
output
[1, 5, 5, 3, 2]
[2, 6, 2, 4, 1]
[3, 9, 1, 6, 1]
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