我正在开发一个 2D 程序生成的 Unity 游戏，我想知道如何在四个基本方向（N、E、S、W）以及四个基本方向（NE 、东南、西南、西北）。
我想要实现的示例：
【问题讨论】：
如果我们将单元格坐标视为row 和column，您可以通过查看上一行、同一行和下一行、前一列、同一列和我们正在搜索的单元格后面的列。
要获得这些值，我们只需设置 minRow = cell.Row - 1、maxRow = cell.Row + 1、minCol = cell.Col - 1 和 maxCol = cell.Col + 1（当然，我们必须检查网格的边界以确保我们不会离开任何边，并且我们不返回 Row 和 Col 与我们正在检查的单元格相同的单元格），并且我们返回具有这些坐标的网格中的所有项目。
例如：
private static List<T> GetNeighbors<T>(int cellRow, int cellCol, T[,] grid)
{
var minRow = cellRow == 0 ? 0 : cellRow - 1;
var maxRow = cellRow == grid.GetUpperBound(0) ? cellRow : cellRow + 1;
var minCol = cellCol == 0 ? 0 : cellCol - 1;
var maxCol = cellCol == grid.GetUpperBound(1) ? cellCol : cellCol + 1;
var results = new List<T>();
for (int row = minRow; row <= maxRow; row++)
{
for (int col = minCol; col <= maxCol; col++)
{
if (row == cellRow && col == cellCol) continue;
results.Add(grid[row, col]);
}
}
return results;
}
实际上，它可能看起来像：
private static void Main()
{
var grid = GetSquareGrid(10);
var neighbors = GetNeighbors(4, 5, grid);
Console.Write($"Neighbors of [4,5] are: ");
Console.Write(string.Join(",", neighbors.Select(n => $"[{n.X},{n.Y}]")));
GetKeyFromUser("\n\nDone! Press any key to exit...");
}
private static Point[,] GetSquareGrid(int size)
{
var result = new Point[size, size];
for (int row = 0; row < size; row++)
{
for (int col = 0; col < size; col++)
{
result[row, col] = new Point(row, col);
}
}
return result;
}
输出：
【讨论】：
所以你有一个二维数组：
var map = new int[20, 20];
首先我们需要宽度和高度（假设您的数组以map[x, y] 的形式排列数据）：
// + 1 because GetUpperBound returns the highest addressable index, not the length in that dimension
var width = map.GetUpperBound(0) + 1;
var height = map.GetUpperBound(1) + 1;
现在我们有了这个，我们可以找到邻居了。我只做过西部、西北、东部和东南部，但我相信你明白了：
int searchX = 5;
int searchY = 5;
// I'm using -1 as a "doesn't exist" value, although you could also look at using nullable int (i.e. int? westNeighbour = null)
int westNeighbour = -1;
int northwestNeighbour = -1;
int eastNeighbour = -1;
int southeastNeighbour = -1;
if (searchX - 1 >= 0) // would this neighbour be on the map?
{
westNeighbour = map[searchX - 1, searchY];
}
if (searchX - 1 >= 0 && searchY - 1 >= 0) // would this neighbour be on the map?
{
northwestNeighbour = map[searchX - 1, searchY - 1];
}
if (searchX + 1 < width) // would this neighbour be on the map?
{
eastNeighbour = map[searchX + 1, searchY];
}
if (searchX + 1 < width && searchY + 1 < height) // would this neighbour be on the map?
{
southeastNeighbour = map[searchX + 1, searchY + 1];
}
请注意，我们正在检查新位置是否仍在数组的边界内，以免出现索引错误。
【讨论】：
假设您使用的是统一的 tilemap，这就是您的操作方式
public static List<TileBase> GetNeighbours(Tilemap tilemap, Vector3Int original)
{
var tiles = new List<TileBase>();
for (int x=-1;x<=1;++x)
{
for (int y=-1;y<=1;++y)
{
var point = new Vector3Int(original.x + x, original.y + y, 0);
if (
tilemap.cellBounds.Contains(point) &&
x!=0 || y!=0
)
{
tiles.Add(tilemap.GetTile(point));
}
}
}
return tiles;
}
【讨论】：
您在评论中提到您正在尝试使用 2D int 数组执行此操作，所以我假设您的图块标有整数。假设您有一个包含所有整数的数组 int[,]。我还将假设您知道要查找邻居的图块的 x,y 位置。
假设您的数组看起来像这样，带有标记的索引：
[0,0] [1,0] [2,0] [3,0]
[0,1] [1,1] [2,1] [3,1]
[0,2] [1,2] [2,2] [3,2]
[0,3] [1,3] [2,3] [3,3]
如果你的数组被翻转，你需要翻转下面的一些逻辑，但它仍然适用。
要让瓷砖向西，它是 x-1。为了让瓷砖向东，它是 x+1。需要注意的是 x 不能大于 2D 数组的宽度，我们将其标记为 int widthOfArray。南北也一样，限制为 heightOfArray
让我们把它付诸实践：
//Assumption: [x,y] is the current array position that you want to find neighbors
//east and west are going to dictate your x index for finding the neighbor.
westIndex = Mathf.Clamp(x - 1, 0f, widthOfArray);
eastIndex = Mathf.Clamp(x + 1, 0f, widthOfArray);
//north and south are going to dictate your y index for finding the neighbor.
northIndex = Mathf.Clamp(y - 1, 0f, heightOfArray);
southIndex = Mathf.Clamp(y + 1, 0f, heightOfArray);
int northPoint = array[x, northIndex];
int northEastPoint = array[eastIndex, northIndex];
int eastPoint = array[eastIndex, y];
int southEastPoint = array[eastIndex, southIndex];
int southPoint = array[x, southIndex];
//...etc
【讨论】：