将一个锯齿状数组复制到另一个锯齿状数组之上

发布于 2024-08-27 04:19:57 字数 449 浏览 8 评论 0原文

我怎样才能完成将一个锯齿状数组复制到另一个锯齿状数组？例如，我有一个 5x7 数组：

0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0

和一个 4x3 数组：

0,1,1,0
1,1,1,1
0,1,1,0

我希望能够在全零数组上指定特定的起点，例如 (1,1)，并将第二个数组复制到其顶部我会得到这样的结果：

0, 0, 0, 0, 0, 0, 0
0, 0, 1, 1, 0, 0, 0
0, 1, 1, 1, 1, 0, 0
0, 0, 1, 1, 0, 0, 0
0, 0, 0, 0, 0, 0, 0

最好的方法是什么？

原文

How could I accomplish copying one jagged array to another? For instance, I have a 5x7 array of:

0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0

and a 4x3 array of:

0,1,1,0
1,1,1,1
0,1,1,0

I would like to be able to specify a specific start point such as (1,1) on my all zero array, and copy my second array ontop of it so I would have a result such as:

0, 0, 0, 0, 0, 0, 0
0, 0, 1, 1, 0, 0, 0
0, 1, 1, 1, 1, 0, 0
0, 0, 1, 1, 0, 0, 0
0, 0, 0, 0, 0, 0, 0

What would be the best way to do this?

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淡忘如思 2024-09-03 04:19:57

由于示例的平方性质，这似乎更适合二维数组而不是锯齿状数组。但无论哪种方式，你当然可以用老式的方式来完成它并循环它。像（未经测试）

for (int i = 0; i < secondArray.Length; i++)
{
    for (int j = 0; j < secondArray[0].Length; j++)
    {
        firstArray[startingRow + i][startingColumn + j] = secondArray[i][j];
    }
}

编辑：像马克一样，我也有一些轻微的改进，略有不同，但大致相同。

for (int i = 0; i < secondArray.Length; i++)
{
    secondArray[i].CopyTo(firstArray[startingRow + i], startingColumn);
}

Due to the squared nature of your example, this seems more fitting of a 2D array instead of jagged. But either way, you could certainly do it the old fashioned way and loop over it. Something like (untested)

for (int i = 0; i < secondArray.Length; i++)
{
    for (int j = 0; j < secondArray[0].Length; j++)
    {
        firstArray[startingRow + i][startingColumn + j] = secondArray[i][j];
    }
}

Edit: Like Mark, I also had a slight improvement, slightly different but along the same lines.

for (int i = 0; i < secondArray.Length; i++)
{
    secondArray[i].CopyTo(firstArray[startingRow + i], startingColumn);
}

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趁微风不噪 2024-09-03 04:19:57

即使您的输入不是矩形，这也应该起作用：

void copy(int[][] source, int[][] destination, int startRow, int startCol)
{
    for (int i = 0; i < source.Length; ++i)
    {
        int[] row = source[i];
        Array.Copy(row, 0, destination[i + startRow], startCol, row.Length);
    }
}

This should work even if your inputs are not rectangular:

void copy(int[][] source, int[][] destination, int startRow, int startCol)
{
    for (int i = 0; i < source.Length; ++i)
    {
        int[] row = source[i];
        Array.Copy(row, 0, destination[i + startRow], startCol, row.Length);
    }
}

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