在数组 c# 中查找与数字最远和最接近的数字

Question

我有一个练习，我需要找到最远和最接近某个数字（平均值）的数字。有人可以帮助我吗？这是我的代码：Console.WriteLine("Geef 10 gehele positieve getallen.");Console.WriteLine("Getal 1:");int g1 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 2:");int g2 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 3:");int g3 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 4:");int g4 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 5:");int g5 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 6:");int g6 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 7:");int g7 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 8:");int g8 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 9:");int g9 = Convert.ToInt32(Console.ReadLine());Console.WriteLine("Getal 10:");int g10 = Convert.ToInt32(Console.ReadLine());int gemiddelde = (g1 + g2 + g3 + g4 + g5 + g6 + g7 + g8 + g9 + g10)/10;int[] array = new int[10] {g1,g2,g3,g4,g5,g6,g7,g8,g9,g10};

Answer 1

如果你可以使用List<int>而不是数组，它会让事情更容易编码，并且可能更清洁，这取决于你问的是谁。

假设你改变了这一点：

int[] array = new int[10] {g1,g2,g3,g4,g5,g6,g7,g8,g9,g10};

变成这样的列表：

List<int> values = new List<int>(){g1,g2,g3,g4,g5,g6,g7,g8,g9,g10};

聚合将测试每个元素，直到完全测试列表。所以我们可以尝试像这样获得最接近和最远的值

// value you want to check is the average of the list.

// Average is already implemented for the List

var value = values.Average();

// will keep the closest value in the collection to our value we are looking for

// and keep testing with the following value to end up with the final closest

var closestTo = values.Aggregate((x, y) => Math.Abs(x - value) < Math.Abs(y - value) ? x : y);

// same logic as closest except we keep the furthest

var furthestTo = values.Aggregate((x, y) => Math.Abs(x - value) > Math.Abs(y - value) ? x : y);