SortedSet<> .Contains()如何实现自己的比较?

时间:2014-03-27 18:37:12

标签: c# sortedset

我想检查SortedSet<>中是否存在具有给定值的Object,但我不明白自定义比较是如何工作的。在List<>.Exists()我可以使用lambda,但是我不能那样做,而且我没有得到整个界面的东西,而msdn说我需要覆盖int返回函数。

public class Node
{
    public int X, Y;
    public int rand;

    public Node(int x, int y, int r)
    { X = x; Y = y; rand = r; }
}

class Program
{
    static void Main(string[] args)
    {
        SortedSet<Node> mySet = new SortedSet<Node>();
        mySet.Add(new Node(1, 2, 90));
        Node myNode = new Node(1, 2, 50);
        // I want this to check if X and Y are the same
        if (mySet.Contains(myNode, interfaceThing))
            Console.WriteLine("Sth is already on that (X, Y) position");      
    }
}

有没有简单的方法呢?

2 个答案:

答案 0 :(得分:5)

您有两个选项,创建一个实现IComparer<Node>的类(您也应该IEqualityComparer<Node>)并将其传递给有序集的构造函数。

public class NodeComparer : IComparer<Node>, IEqualityComparer<Node>
{
    public int Compare(Node node1, Node node2)
    {
        //Sorts by X then by Y

        //perform the X comparison
        var result = node1.X.CompareTo(node2.X);
        if (result != 0)
            return result;

        //Perform the Y Comparison
        return node1.Y.CompareTo(node2.Y);
    }

    public bool Equals(Node x, Node y)
    {
        if (ReferenceEquals(x, y)) return true;
        if (ReferenceEquals(x, null)) return false;
        if (ReferenceEquals(y, null)) return false;
        if (x.GetType() != y.GetType()) return false;
        return x.X == y.X && x.Y == y.Y && x.rand == y.rand;
    }

    public int GetHashCode(Node obj)
    {
        unchecked
        {
            var hashCode = obj.X;
            hashCode = (hashCode * 397) ^ obj.Y;
            hashCode = (hashCode * 397) ^ obj.rand;
            return hashCode;
        }
    }
}

public class Node
{
    public int X, Y;
    public int rand;

    public Node(int x, int y, int r)
    { X = x; Y = y; rand = r; }
}

class Program
{
    static void Main(string[] args)
    {
        SortedSet<Node> mySet = new SortedSet<Node>(new NodeComparer());
        mySet.Add(new Node(1, 2, 90));
        Node myNode = new Node(1, 2, 50);
        // I want this to check if X and Y are the same
        if (mySet.Contains(myNode, interfaceThing))
            Console.WriteLine("Sth is already on that (X, Y) position");      
    }
}

或让Node实现自己需要的相关方法。

public class Node : IEquatable<Node>, IComparable<Node>
{
    public int X, Y;

    public int rand;

    public Node(int x, int y, int r)
    { X = x; Y = y; rand = r; }

    public bool Equals(Node other)
    {
        if (ReferenceEquals(null, other)) return false;
        if (ReferenceEquals(this, other)) return true;
        return X == other.X && Y == other.Y && rand == other.rand;
    }

    public override bool Equals(object obj)
    {
        if (ReferenceEquals(null, obj)) return false;
        if (ReferenceEquals(this, obj)) return true;
        if (obj.GetType() != this.GetType()) return false;
        return Equals((Node)obj);
    }

    public override int GetHashCode()
    {
        unchecked
        {
            var hashCode = X;
            hashCode = (hashCode*397) ^ Y;
            hashCode = (hashCode*397) ^ rand;
            return hashCode;
        }
    }

    public int CompareTo(Node other)
    {
        //First order by X then order by Y then order by rand

        var result = X.CompareTo(other.X);
        if (result != 0)
            return result;

        result = Y.CompareTo(other.Y);
        if (result != 0)
            return result;

        return rand.CompareTo(other.rand);
    }
}

答案 1 :(得分:0)

一个简单的脏方法是使用一些linq

if(myNode.Where(n => n.X == myNode.X && n.Y == myNode.Y).Count > 0)

您也可以在扩展方法中执行此操作,以便能够多次调用

public static class Extensions
{
    public static bool ContainsNode(this IList<Node> nodes, Node value)
    {
        return nodes.Where(n => n.X == value.X && n.Y == value.Y).Count > 0;
    }
}

虽然如果你想提高效率,你应该使用一个简单的foreach循环,一旦找到一个循环,可能会快速遍历整个列表。

修改:完全忘记了.Any()贝司手为你做了什么,但确实早早就切断了。

public static class Extensions
{
    public static bool ContainsNode(this IList<Node> nodes, Node value)
    {
        return nodes.Any(n => n.X == value.X && n.Y == value.Y).Count > 0;
    }
}