Listing all permutations of a string/integer

Question

A common task in programming interviews (not from my experience of interviews though) is to take a string or an integer and list every possible permutation.

Is there

Answer 1

Here is an easy to understand permutaion function for both string and integer as input. With this you can even set your output length(which in normal case it is equal to input length)

String

    static ICollection result;

    public static ICollection GetAllPermutations(string str, int outputLength)
    {
        result = new List();
        MakePermutations(str.ToCharArray(), string.Empty, outputLength);
        return result;
    }

    private static void MakePermutations(
       char[] possibleArray,//all chars extracted from input
       string permutation,
       int outputLength//the length of output)
    {
         if (permutation.Length < outputLength)
         {
             for (int i = 0; i < possibleArray.Length; i++)
             {
                 var tempList = possibleArray.ToList();
                 tempList.RemoveAt(i);
                 MakePermutations(tempList.ToArray(), 
                      string.Concat(permutation, possibleArray[i]), outputLength);
             }
         }
         else if (!result.Contains(permutation))
            result.Add(permutation);
    }

and for Integer just change the caller method and MakePermutations() remains untouched:

    public static ICollection GetAllPermutations(int input, int outputLength)
    {
        result = new List();
        MakePermutations(input.ToString().ToCharArray(), string.Empty, outputLength);
        return result.Select(m => int.Parse(m)).ToList();
    }

example 1: GetAllPermutations("abc",3); "abc" "acb" "bac" "bca" "cab" "cba"

example 2: GetAllPermutations("abcd",2); "ab" "ac" "ad" "ba" "bc" "bd" "ca" "cb" "cd" "da" "db" "dc"

example 3: GetAllPermutations(486,2); 48 46 84 86 64 68