使用HashSet: list = new HashSet<T>(list).ToList();

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)

使用HashSet可以很容易地做到这一点。

List<int> listWithDuplicates = new List<int> { 1, 2, 1, 2, 3, 4, 5 };
HashSet<int> hashWithoutDuplicates = new HashSet<int> ( listWithDuplicates );
List<int> listWithoutDuplicates = hashWithoutDuplicates.ToList();

正如kronoz在. net 3.5中所说，您可以使用Distinct()。

在。net 2中，你可以模仿它:

public IEnumerable<T> DedupCollection<T> (IEnumerable<T> input) 
{
    var passedValues = new HashSet<T>();

    // Relatively simple dupe check alg used as example
    foreach(T item in input)
        if(passedValues.Add(item)) // True if item is new
            yield return item;
}

这可用于删除任何集合，并将以原始顺序返回值。

通常，过滤一个集合(Distinct()和这个示例都是这样做的)比从其中删除项要快得多。

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 }
 */