在。net 2.0中还有另一种方法

    static void Main(string[] args)
    {
        List<string> alpha = new List<string>();

        for(char a = 'a'; a <= 'd'; a++)
        {
            alpha.Add(a.ToString());
            alpha.Add(a.ToString());
        }

        Console.WriteLine("Data :");
        alpha.ForEach(delegate(string t) { Console.WriteLine(t); });

        alpha.ForEach(delegate (string v)
                          {
                              if (alpha.FindAll(delegate(string t) { return t == v; }).Count > 1)
                                  alpha.Remove(v);
                          });

        Console.WriteLine("Unique Result :");
        alpha.ForEach(delegate(string t) { Console.WriteLine(t);});
        Console.ReadKey();
    }

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 }
 */

在Java中(我认为c#或多或少是相同的):

list = new ArrayList<T>(new HashSet<T>(list))

如果你真的想改变原来的列表:

List<T> noDupes = new ArrayList<T>(new HashSet<T>(list));
list.clear();
list.addAll(noDupes);

为了保持顺序，只需将HashSet替换为LinkedHashSet。

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

这里有一个简单的解决方案，不需要任何难读的LINQ或任何列表的预先排序。

   private static void CheckForDuplicateItems(List<string> items)
    {
        if (items == null ||
            items.Count == 0)
            return;

        for (int outerIndex = 0; outerIndex < items.Count; outerIndex++)
        {
            for (int innerIndex = 0; innerIndex < items.Count; innerIndex++)
            {
                if (innerIndex == outerIndex) continue;
                if (items[outerIndex].Equals(items[innerIndex]))
                {
                    // Duplicate Found
                }
            }
        }
    }

可能更简单的方法是确保没有将重复项添加到列表中。

if(items.IndexOf(new_item) < 0) 
    items.add(new_item)