Print repeated numbers only one time in c#

I have an array with "n" numbers and I need to print开发者_如何转开发 all repeated numbers only one time

i made this code, but something is wrong

for (int i = 0; i < numbers.Length; i++)
            {
                for (int j = 1; j < numbers.Length; j++)
                {
                    if (numbers[i] == numbers[j] && i!=j)
                    {
                        Console.WriteLine(numbers[i]);
                        break;
                    }
                }
            }

then if my array have the elements {2,3,1,5,2,3}

the program prints :

2
3
3

what can I do?

You can use:

using System.Linq;

…

foreach(var number in numbers.Distinct()) Console.WriteLine(number);

edit

I may have misunderstood the requirement. If you only want to output the numbers that appear more than once, then you can use:

foreach(var group in numbers.GroupBy(n => n).Where(g => g.Count() > 1))
    Console.WriteLine(group.Key);

var query = numbers.GroupBy(x => x)
                   .Where(g => g.Skip(1).Any())
                   .Select(g => g.Key);

foreach (int n in query)
{
    Console.WriteLine(n);
}

Or, alternatively...

var dict = new Dictionary<int, int>();
foreach (int n in numbers)
{
    int count;
    dict.TryGetValue(n, out count);
    if (count == 1)
    {
        Console.WriteLine(n);
    }
    dict[n] = count + 1;
}

The problem in your code: you get 3 repeated because when i is 1 (looking at the first 3) there's another 3 in the list at the end, and when i is 5 (looking at the last 3) there's another three in the list near the beginning.

Instead you should look at only those numbers which come after your current position - change to int j = i; so that you only look at the positions after your current position, and you won't get repeated results.

for (int i = 0; i < numbers.Length; i++)
{
    for (int j = i; j < numbers.Length; j++)
    {
        if (numbers[i] == numbers[j] && i!=j)
        {
            Console.WriteLine(numbers[i]);
            break;
        }
    }
}

Having said that, your algorithm is not as efficient as using a built in algorithm. Try GroupBy

var duplicates = numbers.GroupBy(n => n)
    .Where(group => group.Count() > 1);

foreach (var group in duplicates)
{
    Console.WriteLine("{0} appears {1} times", group.Key, group.Count());
}

One way to get distinct numbers is

var uniqueNumbers = numbers.Distinct().ToArray()

and then iterate over uniqueNumbers like you have in your snippet with numbers.

You can add the number to a HashSet as you loop and then only print if not in hash set.

That allows you to do it in one pass.

The algorithm above is n^2 which should be avoided.

// deletes an integer if it appears double
#include <iostream.h>
#include <conio.h>

int main ()
{
    int count=0;
   int ar[10]={1,2,3,3,3,4,5,6,7,7};
 for (int i=0; i<10; i++)
  {

                     if (ar[i]==ar[i+1])
                     count++;
                     else
                     cout << ar[i];


}

getch();
  return 0;
}