private static string[] distinct(string[] inputArray)
        {
            bool alreadyExists;
            string[] outputArray = new string[] {};

            for (int i = 0; i < inputArray.Length; i++)
            {
                alreadyExists = false;
                for (int j = 0; j < outputArray.Length; j++)
                {
                    if (inputArray[i] == outputArray[j])
                        alreadyExists = true;
                }
                        if (alreadyExists==false)
                        {
                            Array.Resize<string>(ref outputArray, outputArray.Length + 1);
                            outputArray[outputArray.Length-1] = inputArray[i];
                        }
            }
            return outputArray;
        }

using System;
using System.Collections.Generic;
using System.Linq;


namespace Rextester
{
    public class Program
    {
        public static void Main(string[] args)
        {
             List<int> listofint1 = new List<int> { 4, 8, 4, 1, 1, 4, 8 };
           List<int> updatedlist= removeduplicate(listofint1);
            foreach(int num in updatedlist)
               Console.WriteLine(num);
        }


        public static List<int> removeduplicate(List<int> listofint)
         {
             List<int> listofintwithoutduplicate= new List<int>();


              foreach(var num in listofint)
                 {
                  if(!listofintwithoutduplicate.Any(p=>p==num))
                        {
                          listofintwithoutduplicate.Add(num);
                        }
                  }
             return listofintwithoutduplicate;
         }
    }



}

你可以使用一个LINQ查询来做到这一点:

int[] s = { 1, 2, 3, 3, 4};
int[] q = s.Distinct().ToArray();

泛型扩展方法:

public static IEnumerable<TSource> Distinct<TSource>(this IEnumerable<TSource> source, IEqualityComparer<TSource> comparer)
{
    if (source == null)
        throw new ArgumentNullException(nameof(source));

    HashSet<TSource> set = new HashSet<TSource>(comparer);
    foreach (TSource item in source)
    {
        if (set.Add(item))
        {
            yield return item;
        }
    }
}

下面是一个简单的java逻辑，你遍历数组的元素两次，如果你看到任何相同的元素，你赋0给它，加上你不触及你正在比较的元素的索引。

import java.util.*;
class removeDuplicate{
int [] y ;

public removeDuplicate(int[] array){
    y=array;

    for(int b=0;b<y.length;b++){
        int temp = y[b];
        for(int v=0;v<y.length;v++){
            if( b!=v && temp==y[v]){
                y[v]=0;
            }
        }
    }
}

注意:未测试!

string[] test(string[] myStringArray)
{
    List<String> myStringList = new List<string>();
    foreach (string s in myStringArray)
    {
        if (!myStringList.Contains(s))
        {
            myStringList.Add(s);
        }
    }
    return myStringList.ToString();
}

也许能满足你的需要…

编辑啊! !不到一分钟就被抢了!