首页 > 解决方案 > 如何获取从固定点到特定距离的坐标邻居

问题描述

我的任务是实现 GetNeighbors 方法,该方法从一组点返回仅是具有整数坐标 x 和 y 的点的 h 邻居的点。(阅读代码中的摘要以获得更多理解)

实际上,我和这个人有同样的任务Find k 最近邻居 in C#,但那里的答案似乎不起作用。而且我想只用我猜应该期望的循环来做这件事:)>。

现在我的代码到目前为止:

        /// <summary>
        /// Gets from a set of points only points that are h-neighbors for a point with integer coordinates x and y.
        /// </summary>
        /// <param name="point">Given point with integer coordinates x and y.</param>
        /// <param name="h">Distance around a given point.</param>
        /// <param name="points">A given set of points.</param>
        /// <returns>Only points that are h-neighbors for a point with integer coordinates x and y.</returns>
        /// <exception cref="ArgumentNullException">Throw when array points is null.</exception>
        /// <exception cref="ArgumentException">Throw when h-distance is less or equals zero.</exception>
        public static Point[] GetNeighbors(Point point, int h, params Point[] points)
        {
            if (points is null)
            {
                throw new ArgumentNullException(nameof(points));
            }

            if (h <= 0)
            {
                throw new ArgumentException(null);
            }

            List<Point> neighbors = new List<Point>();

            int left = point.X - h;
            int right = point.X + h;
            int bottom = point.Y - h;
            int top = point.Y + h;


            for (int y = top; y <= bottom; y++)
            {
                for (int x = left; x <= right; x++)
                {
                    // Yeah...
                }
            }

            return neighbors.ToArray();
        }

所以到目前为止我所做的是在邻域中找到顶部、底部、左侧和右侧边界。我认为我只需要做一个 if 语句,它可以比较我拥有的点和来自 points 数组的点。嗯如何做到这一点,我对使用结构不是很熟悉,每次都是失败的。

这就是 struct Point 的构建方式:

/// <summary>
    /// Represents a point on the coordinate plane.
    /// </summary>
    public readonly struct Point : System.IEquatable<Point>
    {
        public Point(int x, int y)
        {
            this.X = x;
            this.Y = y;
        }

        public int X { get; }

        public int Y { get; }
        
        public static bool operator ==(Point left, Point right)
        {
            return left.Equals(right);
        }
        
        public static bool operator !=(Point left, Point right)
        {
            return !(left == right);
        }
        
        public override int GetHashCode()
        {
            return this.X.GetHashCode() ^ this.Y.GetHashCode();
        }
        
        public override bool Equals(object obj)
        {
            if (obj is null)
            {
                return false;
            }

            if (!(obj is Point))
            {
                return false;
            }

            if (obj.GetType() != this.GetType())
            {
                return false;
            }

            Point point = (Point)obj;
            
            return this.Equals(point);
        }

        public bool Equals(Point other)
        {
            return this.X == other.X && this.Y == other.Y;
        }
    }

邻里

标签: c#arraysalgorithmstructcoordinates

解决方案


您不能循环遍历坐标,而是遍历点。这是使用 LINQ 的解决方案:

public static Point[] GetNeighbors(Point point, int h, params Point[] points)
{
    int minX = point.X - h;
    int maxX = point.X + h;
    int minY = point.Y - h;
    int maxY = point.Y + h;
    return points
        .Where(p => p.X >= minX && p.X <= maxX && p.Y >= minY && p.Y <= maxY)
        .ToArray();
}

使用 C# 9.0,您可以简化与模式匹配的比较

        .Where(p => p.X is >=minX and <=maxX && p.Y is >=minY and <=maxY)

这是使用显式循环的解决方案。由于我们不知道结果数组的大小,我将结果存储在一个列表中:

    ...
    var list = new List<Point>();
    foreach (Point p in points) {
        if (p.X is >=minX and <=maxX && p.Y is >=minY and <=maxY) {
            list.Add(p);
        }
    }
    return list.ToArray();

或者您可以返回一个列表。


另一种解决方案是使用迭代器方法

public static IEnumerable<Point> GetNeighbors(Point point, int h, params Point[] points)
{
    int minX = point.X - h;
    int maxX = point.X + h;
    int minY = point.Y - h;
    int maxY = point.Y + h;
    foreach (Point p in points) {
        if (p.X is >=minX and <=maxX && p.Y is >=minY and <=maxY) {
            yield return p;
        }
    }
}

请注意,这种方法不会将点存储在集合中,而是在循环结果时执行方法内部的循环(显式地在 foreach 循环中或通过应用等隐式地执行.ToArray().Count()

Point[] result = GetNeighbors(center, 2, p1, p2, p3, p4).ToArray();

推荐阅读