permutation in c#

Question

Is it possible to generate all permutations of a collection in c#?

char[] inputSet = { \'A\',\'B\',\'C\' };
Permutations permutations = new Permu         


        

Answer 1

I've already faced the problem and I wrote these simple methods:

    public static IList GeneratePermutations(T[] objs, long? limit)
    {
        var result = new List();
        long n = Factorial(objs.Length);
        n = (!limit.HasValue || limit.Value > n) ? n : (limit.Value);

        for (long k = 0; k < n; k++)
        {
            T[] kperm = GenerateKthPermutation(k, objs);
            result.Add(kperm);
        }

        return result;
    }

    public static T[] GenerateKthPermutation(long k, T[] objs)
    {
        T[] permutedObjs = new T[objs.Length];

        for (int i = 0; i < objs.Length; i++)
        {
            permutedObjs[i] = objs[i];
        }
        for (int j = 2; j < objs.Length + 1; j++)
        {
            k = k / (j - 1);                      // integer division cuts off the remainder
            long i1 = (k % j);
            long i2 = j - 1;
            if (i1 != i2)
            {
                T tmpObj1 = permutedObjs[i1];
                T tmpObj2 = permutedObjs[i2];
                permutedObjs[i1] = tmpObj2;
                permutedObjs[i2] = tmpObj1;
            }
        }
        return permutedObjs;
    }

    public static long Factorial(int n)
    {
        if (n < 0) { throw new Exception("Unaccepted input for factorial"); }    //error result - undefined
        if (n > 256) { throw new Exception("Input too big for factorial"); }  //error result - input is too big

        if (n == 0) { return 1; }

        // Calculate the factorial iteratively rather than recursively:

        long tempResult = 1;
        for (int i = 1; i <= n; i++)
        {
            tempResult *= i;
        }
        return tempResult;
    }

Usage:

var perms = Utilities.GeneratePermutations(new char[]{'A','B','C'}, null);