Question

基本上，我在C＃中有一个完整的结构列表，看起来像这样（代码可能不合适语法）

public struct Numbers {
    public int DistanceFromArbitraryPoint;
    public int X;
    public int Y;
}

我想做的是对列表进行排序：

按距离排序列表（已经没有问题）。
对于距离<1的每个数字。 SomeCriteria，按X和Y的总和对这些列表项进行排序，而不创建新列表。只需重新排列列表即可。

结果将成为：

Distance 12, x 0, y 0;
Distance 4, x 20, y 20;
Distance 6, x 0, y 3;

第一步之后：

Distance 4, x 20, y 20;
Distance 6, x 0, y 3;
Distance 12, x 0, y 0;

在步骤2之后（如果距离<10，则按X和Y的总和进行排序）

Distance 6, x 0, y 3; // lower X and Y Sum, goes first
Distance 4, x 20, y 20;
Distance 12, x 0, y 0; // Distance higher than 10, remains unsorted

Answer 1

这里需要更复杂的比较方法

var numbers = new List<Numbers>{ new Numbers(6,0,3),   new Numbers(12,0,0), 
                                 new Numbers(4,20,20), new Numbers(5,20,20)};       
int d = 10;

numbers.Sort((a,b)=>
             {
                 if (a.DistanceFromArbitraryPoint >= d && b.DistanceFromArbitraryPoint >= d)                         
                     return a.DistanceFromArbitraryPoint.CompareTo(b.DistanceFromArbitraryPoint);                        
                 if (a.DistanceFromArbitraryPoint >= d)
                     return 1;
                 if (b.DistanceFromArbitraryPoint >= d)
                     return -1;

                 int c = (a.X+a.Y).CompareTo(b.X+b.Y);                       
                 if (c == 0)
                     c = a.DistanceFromArbitraryPoint.CompareTo(b.DistanceFromArbitraryPoint);
                 return c;                       
             });

输出：

6 (0,3); 4 (20,20); 5 (20,20); 12 (0,0)

使用来自Sort(Comparison<T> comparison)的

List<T>，而不是Linq

fiddle

public struct Numbers
{
    public Numbers(int d, int x, int y)
    {
        DistanceFromArbitraryPoint  = d;
        X = x;
        Y = y;
    }
    public int DistanceFromArbitraryPoint;

    public int X;
    public int Y;

    public override string ToString()
    {
        return String.Format("{0} ({1},{2})", DistanceFromArbitraryPoint, X,Y);
    }
}

C＃按列表排序满足条件的项目

1 个答案: