David J。的答案是一个很好的方法，不需要额外的对象，排序等。但是，它可以在以下方面进行改进:

for （int innerIndex = items.计数 - 1;内索引 > 外索引 ;内部索引--）

因此，对于整个列表，外部循环会从上到下，但内部循环会从下到“直到到达外部循环的位置”。

外部循环确保整个列表被处理，内部循环找到实际的重复项，这些只会发生在外部循环还没有处理的部分。

或者如果你不想对内循环做自底向上你可以让内循环从outerIndex + 1开始。

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;
    }
}

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。

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
                        });
                    }
                }

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。

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 }
 */