c# 已知一个list包含一个最大和最小的两个数， 传递一个数与值比较 放回新的最大和最小值，排除0和异常数

在C#中，你可以创建一个名为`FindMinMaxExcludingZeroAndExceptionalNumbers`的方法来实现这个功能。首先，你需要遍历列表，找到当前的最大值和最小值，同时排除0和异常数。下面是一个示例代码：

using System;
using System.Collections.Generic;

public class MaxMinCalculator
{
    public (int max, int min) FindMinMaxExcludingZeroAndExceptionalNumbers(List<int> numbers, int value)
    {
        if (numbers == null || numbers.Count == 0)
            throw new ArgumentException("List is empty or null");

        int? currentMax = null;
        int? currentMin = null;
        
        foreach (int number in numbers)
        {
            if (number == 0)
                continue; // 排除0

            try
            {
                if (currentMax == null || number > currentMax.Value)
                    currentMax = number;
                else if (currentMin == null || number < currentMin.Value)
                    currentMin = number;
            }
            catch (Exception ex) when (ex is ArgumentException || ex is DivideByZeroException) // 排除异常数值
            {
                continue;
            }
        }

        // 如果value不在范围内，则更新max和min
        if (value != 0 && (currentMax == null || value > currentMax.Value))
            currentMax = value;
        if (value != 0 && (currentMin == null || value < currentMin.Value))
            currentMin = value;

        return (currentMax ?? default(int), currentMin ?? default(int));
    }
}

然后，你可以这样使用这个类：

List<int> numbers = new List<int> { 5, -3, 0, 7, 2, 9 };
int valueToCompare = 4;
MaxMinCalculator calculator = new MaxMinCalculator();
var result = calculator.FindMinMaxExcludingZeroAndExceptionalNumbers(numbers, valueToCompare);
Console.WriteLine($"Max: {result.max}, Min: {result.min}");