SortedSet 。包含（）如何实现自己的比较？

我想检查SortedSet是否存在具有给定值的Object，但我不明白自定义比较在这里是如何工作的。 在List.Exists()我可以只使用lambda，但我不能那样做，我不会得到整个接口的东西，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 mySet = new SortedSet(); 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？

Answer 1

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

 public class NodeComparer : IComparer, IEqualityComparer { 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 xX == yX && xY == yY && 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 mySet = new SortedSet(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"); } } 

或者让Node实现自己需要的相关方法。

 public class Node : IEquatable, IComparable { 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); } } 

Answer 2

一个简单的脏方法是使用一些linq

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

您也可以在扩展方法中执行此操作，以便能够多次调用此方法

 public static class Extensions { public static bool ContainsNode(this IList nodes, Node value) { return nodes.Where(n => nX == value.X && nY == 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。

编辑 ：完全忘记了.Any .Any()贝司手为你做了哪里，但确实早了。

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