我有一个二维数组，说

0 0 0 0 0
0 2 3 4 0
0 9 1 5 0
0 8 7 6 0
0 0 0 0 0

和我需要得到所有相邻的编号，1（2，3，4，5，6，7，8，9）

And i need to get all the numbers adjacent to 1(2, 3, 4, 5, 6, 7, 8, 9)

有一个不太难看的解决方案比：

Is there a less ugly solution than:

topLeft = array[x-1][y-1]
top  = array[x][y-1]
topRight = array[x+1][y-1]
# etc

谢谢！

推荐答案

如果你不担心顺序，最干净的可能是使用了几个循环的：

If you're not worried about the order, the cleanest is probably to use a couple of loops:

result = new List<int>(8);
for (dx = -1; dx <= 1; ++dx) {
    for (dy = -1; dy <= 1; ++dy) {
        if (dx != 0 || dy != 0) {
            result.Add(array[x + dx][y + dy]);
        }
    }
}

如果顺序很重要，你可以构造的所有（DX，DY）的列表中，你想要的顺序和迭代的吧。

If the order is important, you can construct a list of all the (dx, dy) in the order you want and iterate over that instead.

正如在评论中，你可能要添加边界检查。你可以做这样的（假设顺序并不重要）：

As pointed out in the comments, you probably want to add boundary checks. You could do that like this (assuming order doesn't matter):

List<int> result = new List<int>(8);
for (int dx = (x > 0 ? -1 : 0); dx <= (x < max_x ? 1 : 0); ++dx)
{
    for (int dy = (y > 0 ? -1 : 0); dy <= (y < max_y ? 1 : 0); ++dy)
    {
        if (dx != 0 || dy != 0)
        {
            result.Add(array[x + dx][y + dy]);
        }
    }
}