连接多个数组的简单代码:

string[] arr1 = ...
string[] arr2 = ...
string[] arr3 = ...    
List<string> arr = new List<string>(arr1.Length + arr2.Length + arr3.Length);
arr.AddRange(arr1);
arr.AddRange(arr2);
arr.AddRange(arr3);
string[] result = arr.ToArray();

string[] names1 = new string[] { "Ava", "Emma", "Olivia" };
string[] names2 = new string[] { "Olivia", "Sophia", "Emma" };
List<string> arr = new List<string>(names1.Length + names2.Length);
arr.AddRange(names1);
arr.AddRange(names2);
string[] result = arr.Distinct().ToArray();
foreach(string str in result)
{
    Console.WriteLine(str.ToString());
}

Console.ReadLine();

连接多个数组的简单代码:

string[] arr1 = ...
string[] arr2 = ...
string[] arr3 = ...    
List<string> arr = new List<string>(arr1.Length + arr2.Length + arr3.Length);
arr.AddRange(arr1);
arr.AddRange(arr2);
arr.AddRange(arr3);
string[] result = arr.ToArray();

如果你可以操作其中一个数组，你可以在执行复制之前调整它的大小:

T[] array1 = getOneArray();
T[] array2 = getAnotherArray();
int array1OriginalLength = array1.Length;
Array.Resize<T>(ref array1, array1OriginalLength + array2.Length);
Array.Copy(array2, 0, array1, array1OriginalLength, array2.Length);

否则，您可以创建一个新数组

T[] array1 = getOneArray();
T[] array2 = getAnotherArray();
T[] newArray = new T[array1.Length + array2.Length];
Array.Copy(array1, newArray, array1.Length);
Array.Copy(array2, 0, newArray, array1.Length, array2.Length);

更多关于MSDN上可用的Array方法。

自从。net 5以来，我们现在有了AllocateUnitializedArray，它可能会为建议的解决方案增加额外的(小)性能改进:

public static T[] ConcatArrays<T>(IEnumerable<T[]> arrays)
{
    var result = GC.AllocateUnitializedArray<T>(arrays.Sum(a => a.Length));
    var offset = 0;
    foreach (var a in arrays)
    {
        a.CopyTo(result, offset);
        offset += a.Length;
    }
    return result;
}

创建和扩展方法来处理null

public static class IEnumerableExtenions
{
    public static IEnumerable<T> UnionIfNotNull<T>(this IEnumerable<T> list1, IEnumerable<T> list2)
    {
        if (list1 != null && list2 != null)
            return list1.Union(list2);
        else if (list1 != null)
            return list1;
        else if (list2 != null)
            return list2;
        else return null;
    }
}