decimal[] array = new decimal[5]{80.23,60.20,88.01,77.00,20.45};
decimal TargetNumber = 70.40;
这里,最近的值是77.00,如何找到最接近的十进制数组的索引？
注意：它应该保持与我需要的最接近值的精确索引相同的顺序.在这里,指数比价值重要
解决方法
关于使用LINQ做很多事情的一个好处是,如果找到完全匹配,你可以提前停止检查. minIndex保存索引,如果数组为空,则保持-1.
decimal minDistance = 0; //0 is fine here it is never read,it is just to make the compiler happy.
int minIndex = -1;
for(int i = 0; i < array.Length; i++)
{
var distance = Math.Abs(TargetNumber - array[i]);
if(minIndex == -1 || distance < minDistance)
{
minDistance = distance;
minIndex = i;
//Optional,stop testing if we find a exact match.
if(minDistance == 0)
break;
}
}
为了好玩,我制作了一个完全通用的版本,它要求你传入一个委托来计算距离因子,它还有一个可选参数来定义停止检查更多结果所需的“最小距离”.
using System;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
decimal[] array = new decimal[5]{80.23M,80.40M,80.80M,80.00M,20.45M};
decimal TargetNumber = 70.40M;
var result = FindClosestIndex(TargetNumber,array,(target,element) => Math.Abs(target - element)); //Optionally add in a "(distance) => distance == 0" at the end to enable early termination.
Console.WriteLine(result);
}
public static int FindClosestIndex<T,U>(T target,IEnumerable<T> elements,Func<T,T,U> distanceCalculator,Func<U,bool> earlyTermination = null) where U : IComparable<U>
{
U minDistance = default(U);
int minIndex = -1;
using(var enumerator = elements.GetEnumerator())
for(int i = 0; enumerator.MoveNext(); i++)
{
var distance = distanceCalculator(enumerator.Current,target);
if(minIndex == -1 || minDistance.CompareTo(distance) > 0)
{
minDistance = distance;
minIndex = i;
}
if(earlyTermination != null && earlyTermination(minDistance))
break;
}
return minIndex;
}
}
