本文介绍了获取二维数组中的相邻元素?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个二维数组,比如说
I have a two-dimensional array, say
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
我需要得到与 1(2, 3, 4, 5, 6, 7, 8, 9) 相邻的所有数字
And i need to get all the numbers adjacent to 1(2, 3, 4, 5, 6, 7, 8, 9)
有没有比以下更丑陋的解决方案:
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几个循环:
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,您可以按照您想要的顺序构建所有 (dx, dy) 的列表,然后对其进行迭代.
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.
正如评论中所指出的,您可能想要添加边界检查.你可以这样做(假设顺序无关紧要):
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]);
}
}
}
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