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();

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。

这对我很管用。简单地使用

List<Type> liIDs = liIDs.Distinct().ToList<Type>();

将“类型”替换为所需的类型，例如int。

根据删除重复，我们必须应用下面的逻辑，所以它将以快速的方式删除重复。

public class Program
{

    public static void Main(string[] arges)
    {
        List<string> cities = new List<string>() { "Chennai", "Kolkata", "Mumbai", "Mumbai","Chennai", "Delhi", "Delhi", "Delhi", "Chennai", "Kolkata", "Mumbai", "Chennai" };
        cities = RemoveDuplicate(cities);

        foreach (var city in cities)
        {
            Console.WriteLine(city);
        }
    }

    public static List<string> RemoveDuplicate(List<string> cities)
    {
        if (cities.Count < 2)
        {
            return cities;
        }

        int size = cities.Count;
        for (int i = 0; i < size; i++)
        {
            for (int j = i+1; j < size; j++)
            {
                if (cities[i] == cities[j])
                {
                    cities.RemoveAt(j);
                    size--;
                    j--;
                }
            }
        }
        return cities;
    }
}

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

作为一个辅助方法(没有Linq):

public static List<T> Distinct<T>(this List<T> list)
{
    return (new HashSet<T>(list)).ToList();
}