1 示例

1.1 以PointF为例展示效果

private void button1_Click(object sender, EventArgs e)
{
    Random random = new Random();
    List<PointF> list = new List<PointF>();
    for (int i = 0; i < 10; i++)
    {
        list.Add(new PointF(i, random.Next(1, 100)));
    }

    var listSortMaxToMin = list.ToList();
    listSortMaxToMin.Sort(CompareMaxToMin);

    var listSortMinToMax = list.ToList();
    listSortMaxToMin.Sort(CompareMinToMax);

}

private static int CompareMaxToMin(PointF p1, PointF p2)
{
    if (p1.Y == p2.Y)
        return 0;
    else if (p1.Y < p2.Y)
        return 1;
    else
    {
        return -1;
    }
}

private static int CompareMinToMax(PointF p1, PointF p2)
{
    if (p1.Y == p2.Y)
        return 0;
    else if (p1.Y < p2.Y)
        return 1;
    else
    {
        return -1;
    }
}

1.2 运行结果展示

2实际运用

2.1 创建结构体

public struct MyPoint
{
    public double X;
    public double Y;

    public MyPoint(double x,double y)
    {
        X = x;
        Y = y;
    }
}

2.2 调用示例

/// <summary>
/// 获取(元素相互匹配)两个数组中，arr2元素值最大的count个值对应的的arr1元素
/// </summary>
public void GetMostValue(double[] arr1, double[] arr2, uint count, out List<MyPoint> result)
{
    result = new List<MyPoint>();
    if (arr1 == null) throw new Exception("arrX不可为null");
    if (arr2 == null) throw new Exception("arrY不可为null");
    if (arr1.Length != arr2.Length) throw new Exception("数组arrX和arrY的元素总数不同");
    if (arr1.Length < count) throw new Exception("数组元素数不同不可小于count");

    List<MyPoint> list = new List<MyPoint>();
    int length = arr1.Length;
    for (int i = 0; i < length; i++)
    {
        list.Add(new MyPoint(arr1[i], arr2[i]));
    }

    list.Sort((p1, p2) =>
    {
        if (p1.Y < p2.Y) return 1;
        if (p1.Y > p2.Y) return -1;
        return 0;
    });

    for (int i = 0; i < count; i++)
    {
        result.Add(list[i]);
    }
}

2.3 运行结果展示