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;
    }
}

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 }
 */

有很多方法可以解决列表中的重复问题，下面是其中之一:

List<Container> containerList = LoadContainer();//Assume it has duplicates
List<Container> filteredList = new  List<Container>();
foreach (var container in containerList)
{ 
  Container duplicateContainer = containerList.Find(delegate(Container checkContainer)
  { return (checkContainer.UniqueId == container.UniqueId); });
   //Assume 'UniqueId' is the property of the Container class on which u r making a search

    if(!containerList.Contains(duplicateContainer) //Add object when not found in the new class object
      {
        filteredList.Add(container);
       }
  }

干杯 拉维Ganesan

一个简单直观的实现:

public static List<PointF> RemoveDuplicates(List<PointF> listPoints)
{
    List<PointF> result = new List<PointF>();

    for (int i = 0; i < listPoints.Count; i++)
    {
        if (!result.Contains(listPoints[i]))
            result.Add(listPoints[i]);
        }

        return result;
    }

我认为最简单的方法是:

创建一个新列表并添加唯一的项目。

例子:

        class MyList{
    int id;
    string date;
    string email;
    }
    
    List<MyList> ml = new Mylist();

ml.Add(new MyList(){
id = 1;
date = "2020/09/06";
email = "zarezadeh@gmailcom"
});

ml.Add(new MyList(){
id = 2;
date = "2020/09/01";
email = "zarezadeh@gmailcom"
});

 List<MyList> New_ml = new Mylist();

foreach (var item in ml)
                {
                    if (New_ml.Where(w => w.email == item.email).SingleOrDefault() == null)
                    {
                        New_ml.Add(new MyList()
                        {
                          id = item.id,
     date = item.date,
               email = item.email
                        });
                    }
                }

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