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;
            }
        }
    }
}

在下面找到答案。

class Program
{
    static void Main(string[] args)
    {
        var nums = new int[] { 1, 4, 3, 3, 3, 5, 5, 7, 7, 7, 7, 9, 9, 9 };
        var result = removeDuplicates(nums);
        foreach (var item in result)
        {
            Console.WriteLine(item);
        }
    }
    static int[] removeDuplicates(int[] nums)
    {
        nums = nums.ToList().OrderBy(c => c).ToArray();
        int j = 1;
        int i = 0;
        int stop = 0;
        while (j < nums.Length)
        {
            if (nums[i] != nums[j])
            {
                nums[i + 1] = nums[j];
                stop = i + 2;
                i++;
            }
            j++;
        }
        nums = nums.Take(stop).ToArray();
        return nums;
    }
}

这是基于我刚刚解决的一个测试的一点贡献，可能对这里其他顶级贡献者的改进有所帮助。 以下是我所做的事情:

I used OrderBy which allows me order or sort the items from smallest to the highest using LINQ I then convert it to back to an array and then re-assign it back to the primary datasource So i then initialize j which is my right hand side of the array to be 1 and i which is my left hand side of the array to be 0, i also initialize where i would i to stop to be 0. I used a while loop to increment through the array by going from one position to the other left to right, for each increment the stop position is the current value of i + 2 which i will use later to truncate the duplicates from the array. I then increment by moving from left to right from the if statement and from right to right outside of the if statement until i iterate through the entire values of the array. I then pick from the first element to the stop position which becomes the last i index plus 2. that way i am able to remove all the duplicate items from the int array. which is then reassigned.

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!

如果您需要对它进行排序，那么您可以实现一个也删除重复项的排序。

那就一石二鸟了。

——这是面试中每次都会问的问题。现在我完成了它的编码。

static void Main(string[] args)
{    
            int[] array = new int[] { 4, 8, 4, 1, 1, 4, 8 };            
            int numDups = 0, prevIndex = 0;

            for (int i = 0; i < array.Length; i++)
            {
                bool foundDup = false;
                for (int j = 0; j < i; j++)
                {
                    if (array[i] == array[j])
                    {
                        foundDup = true;
                        numDups++; // Increment means Count for Duplicate found in array.
                        break;
                    }                    
                }

                if (foundDup == false)
                {
                    array[prevIndex] = array[i];
                    prevIndex++;
                }
            }

            // Just Duplicate records replce by zero.
            for (int k = 1; k <= numDups; k++)
            {               
                array[array.Length - k] = '\0';             
            }


            Console.WriteLine("Console program for Remove duplicates from array.");
            Console.Read();
        }

List<String> myStringList = new List<String> (); foreach (myStringArray中的字符串) ｛ 如果(! myStringList.Contains (s)) ｛ myStringList.Add(年代); ｝ ｝

这是O(n²)，这对于一个短列表来说没有什么关系因为它将被塞进一个组合中，但对于一个大的集合来说可能很快就会成为一个问题。