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.

简单的解决方案:

using System.Linq;
...

public static int[] Distinct(int[] handles)
{
    return handles.ToList().Distinct().ToArray();
}

int size = a.Length;
        for (int i = 0; i < size; i++)
        {
            for (int j = i + 1; j < size; j++)
            {
                if (a[i] == a[j])
                {
                    for (int k = j; k < size; k++)
                    {
                        if (k != size - 1)
                        {
                            int temp = a[k];
                            a[k] = a[k + 1];
                            a[k + 1] = temp;

                        }
                    }
                    j--;
                    size--;
                }
            }
        }

这段代码从数组中100%删除重复值[因为我使用了一个[I]].....您可以将其转换为任何OO语言.....：）

for(int i=0;i<size;i++)
{
    for(int j=i+1;j<size;j++)
    {
        if(a[i] == a[j])
        {
            for(int k=j;k<size;k++)
            {
                 a[k]=a[k+1];
            }
            j--;
            size--;
        }
    }

}

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;
            }
        }
    }
}

public static int RemoveDuplicates(ref int[] array)
{
    int size = array.Length;

    // if 0 or 1, return 0 or 1:
    if (size  < 2) {
        return size;
    }

    int current = 0;
    for (int candidate = 1; candidate < size; ++candidate) {
        if (array[current] != array[candidate]) {
            array[++current] = array[candidate];
        }
    }

    // index to count conversion:
    return ++current;
}