Counting duplicate numbers in a list

I have an list:

int list = { 1,1,2,3,4,4,5,7,7,7,10};

Now I need to make a pr开发者_StackOverflowogram which calculates the double numbers. A number is double when the number before it is the same. I hope you understand. So 1 is double, 4 is doubles and we got 2 double in 7,7,7.

---

Here's a solution in LINQ:

var doubles = list.Skip(1)
                  .Where((number, index) => list[index] == number);

This creates another sequence by skipping the first member of the list, and then finds elements from both sequences that have the same index and the same value. It will run in linear time, but only because a list offers O(1) access by index.

---

Here's an approach which is relatively simple, only iterates once over the sequence, and works with any sequence (not just lists):

public IEnumerable<T> FindConsecutiveDuplicates<T>(this IEnumerable<T> source)
{
    using (var iterator = source.GetEnumerator())
    {
        if (!iterator.MoveNext())
        {
            yield break;
        }
        T current = iterator.Current;
        while (iterator.MoveNext())
        {
            if (EqualityComparer<T>.Default.Equals(current, iterator.Current))
            {
                yield return current;
            }
            current = iterator.Current;
        }
    }
}

Here's another one which is even simpler in that it's only a LINQ query, but it uses side-effects in the Where clause, which is nasty:

IEnumerable<int> sequence = ...;

bool first = true;
int current = 0;
var result = sequence.Where(x => {
   bool result = !first && x == current;
   current = x;
   first = false;
   return result;
});

A third alternative, which is somewhat cleaner but uses a SelectConsecutive method which is basically SelectPairs from this answer, but renamed to be slightly clearer :)

IEnumerable<int> sequence = ...;
IEnumerable<int> result = sequence.SelectConsecutive((x, y) => new { x, y })
                                  .Where(z => z.x == z.y);

---

Everyone seems to be trying to find good ways of doing it, so here's a really bad way instead:

List<int> doubles = new List<int>();
Dictionary<int, bool> seenBefore = new Dictionary<int, bool>();

foreach(int i in list)
{
    try
    {
        seenBefore.Add(i, true);
    }
    catch (ArgumentException)
    {
        doubles.Add(i);
    }
}

return doubles;

Please don't do it like that.

---

something like this may work:

list.GroupBy (l => l).Where (l => l.Count () > 1).SelectMany (l => l).Distinct();

EDIT:

the above code doesn't get the result the OP wanted. Here is an edited version that is taking inspiration from Ani's elegant solution below: :)

list.GroupBy(l => l).Select(g=>g.Skip(1)).SelectMany (l => l);

---

An example that (probably) performs better than using Linq, though is arguably less elegant:

for (int i = 1; i < list.Count; i++)
    if (list[i] == list[i - 1])
        doubles.Add(list[i]);

---

Here you go with the answer in c# :)

int[] intarray = new int[] { 1, 1, 2, 3, 4, 4, 5, 7, 7, 7, 10 };

int previousnumber = -1;
List<int> doubleDigits = new List<int>();
for (int i = 0; i < intarray.Length; i++)
{
    if (previousnumber == -1) { previousnumber = intarray[i]; continue; }
    if (intarray[i] == previousnumber)
    {
        if (!doubleDigits.Contains(intarray[i]))
        {
            doubleDigits.Add(intarray[i]);
            //Console.WriteLine("Duplicate int found - " + intarray[i]);
            continue;
        }
    }
    else
    {
        previousnumber = intarray[i];
    }
}

---

You could do this:

list.GroupBy(i => i).Where(g => g.Count() > 1).SelectMany(g => g.Skip(1))

This is a bit like @KJN's answer, except I think it expresses the "doubles" and "two doubles" clause in the question a bit better:

1. group all integers together
2. only interested in those that appear more than once (g.Count() > 1)
3. select a flattened list of the "doubles", being those after the first (g.Skip(1))

PS: We are assuming here, that GroupBy doesn't first sort the list and if it does, that that sort is not negatively influenced by a pre-sorted list...