扩展方法可能是一种不错的方式……就像这样:

public static List<T> Deduplicate<T>(this List<T> listToDeduplicate)
{
    return listToDeduplicate.Distinct().ToList();
}

然后像这样调用，例如:

List<int> myFilteredList = unfilteredList.Deduplicate();

如果需要比较复杂的对象，则需要在Distinct()方法中传递一个compararer对象。

private void GetDistinctItemList(List<MyListItem> _listWithDuplicates)
{
    //It might be a good idea to create MyListItemComparer 
    //elsewhere and cache it for performance.
    List<MyListItem> _listWithoutDuplicates = _listWithDuplicates.Distinct(new MyListItemComparer()).ToList();
        
    //Choose the line below instead, if you have a situation where there is a chance to change the list while Distinct() is running.
    //ToArray() is used to solve "Collection was modified; enumeration operation may not execute" error.
    //List<MyListItem> _listWithoutDuplicates = _listWithDuplicates.ToArray().Distinct(new MyListItemComparer()).ToList();

    return _listWithoutDuplicates;
}

假设你有另外两个类，比如:

public class MyListItemComparer : IEqualityComparer<MyListItem>
{
    public bool Equals(MyListItem x, MyListItem y)
    {
        return x != null 
               && y != null 
               && x.A == y.A 
               && x.B.Equals(y.B); 
               && x.C.ToString().Equals(y.C.ToString());
    }

    public int GetHashCode(MyListItem codeh)
    {
        return codeh.GetHashCode();
    }
}

And:

public class MyListItem
{
    public int A { get; }
    public string B { get; }
    public MyEnum C { get; }

    public MyListItem(int a, string b, MyEnum c)
    {
        A = a;
        B = b;
        C = c;
    }
}

所有的答案要么复制列表，要么创建一个新列表，要么使用慢函数，要么就是慢得令人痛苦。

据我所知，这是我所知道的最快和最便宜的方法(同时，还得到了一个非常有经验的实时物理优化程序员的支持)。

// Duplicates will be noticed after a sort O(nLogn)
list.Sort();

// Store the current and last items. Current item declaration is not really needed, and probably optimized by the compiler, but in case it's not...
int lastItem = -1;
int currItem = -1;

int size = list.Count;

// Store the index pointing to the last item we want to keep in the list
int last = size - 1;

// Travel the items from last to first O(n)
for (int i = last; i >= 0; --i)
{
    currItem = list[i];

    // If this item was the same as the previous one, we don't want it
    if (currItem == lastItem)
    {
        // Overwrite last in current place. It is a swap but we don't need the last
       list[i] = list[last];

        // Reduce the last index, we don't want that one anymore
        last--;
    }

    // A new item, we store it and continue
    else
        lastItem = currItem;
}

// We now have an unsorted list with the duplicates at the end.

// Remove the last items just once
list.RemoveRange(last + 1, size - last - 1);

// Sort again O(n logn)
list.Sort();

最终成本为:

nlogn + n + nlogn = n + 2nlogn = O(nlogn)非常漂亮。

关于RemoveRange注意事项: 由于我们不能设置列表的计数并避免使用Remove函数，我不知道这个操作的确切速度，但我猜这是最快的方法。

如果你不关心顺序，你可以把这些项推到HashSet中，如果你想保持顺序，你可以这样做:

var unique = new List<T>();
var hs = new HashSet<T>();
foreach (T t in list)
    if (hs.Add(t))
        unique.Add(t);

或者用Linq的方式:

var hs = new HashSet<T>();
list.All( x =>  hs.Add(x) );

编辑:HashSet方法是O(N)时间和O(N)空间，而排序，然后使唯一(由@lassevk和其他人建议)是O(N*lgN)时间和O(1)空间，所以我不太清楚(因为它是第一眼)，排序方式是较差的

也许您应该考虑使用HashSet。

从MSDN链接:

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        HashSet<int> evenNumbers = new HashSet<int>();
        HashSet<int> oddNumbers = new HashSet<int>();

        for (int i = 0; i < 5; i++)
        {
            // Populate numbers with just even numbers.
            evenNumbers.Add(i * 2);

            // Populate oddNumbers with just odd numbers.
            oddNumbers.Add((i * 2) + 1);
        }

        Console.Write("evenNumbers contains {0} elements: ", evenNumbers.Count);
        DisplaySet(evenNumbers);

        Console.Write("oddNumbers contains {0} elements: ", oddNumbers.Count);
        DisplaySet(oddNumbers);

        // Create a new HashSet populated with even numbers.
        HashSet<int> numbers = new HashSet<int>(evenNumbers);
        Console.WriteLine("numbers UnionWith oddNumbers...");
        numbers.UnionWith(oddNumbers);

        Console.Write("numbers contains {0} elements: ", numbers.Count);
        DisplaySet(numbers);
    }

    private static void DisplaySet(HashSet<int> set)
    {
        Console.Write("{");
        foreach (int i in set)
        {
            Console.Write(" {0}", i);
        }
        Console.WriteLine(" }");
    }
}

/* This example produces output similar to the following:
 * evenNumbers contains 5 elements: { 0 2 4 6 8 }
 * oddNumbers contains 5 elements: { 1 3 5 7 9 }
 * numbers UnionWith oddNumbers...
 * numbers contains 10 elements: { 0 2 4 6 8 1 3 5 7 9 }
 */

你可以使用Union

obj2 = obj1.Union(obj1).ToList();