Reputation: 15925
After using Google to find examples, I found the following example goes which works for generating permutations:
namespace ConsoleApp1
{
class Program
{
public static void Main()
{
int n, i;
formPermut test = new formPermut();
int[] arr1 = new int[5];
Console.WriteLine("\n\n Recursion : Generate all possible permutations of an array :");
Console.WriteLine("------------------------------------------------------------------");
Console.Write(" Input the number of elements to store in the array [maximum 5 digits ] :");
n = Convert.ToInt32(Console.ReadLine());
Console.Write(" Input {0} number of elements in the array :\n", n);
for (i = 0; i < n; i++)
{
Console.Write(" element - {0} : ", i);
arr1[i] = Convert.ToInt32(Console.ReadLine());
}
Console.Write("\n The Permutations with a combination of {0} digits are : \n", n);
test.prnPermut(arr1, 0, n - 1);
Console.Write("\n\n");
Console.ReadKey();
}
class formPermut
{
public void swapTwoNumber(ref int a, ref int b)
{
int temp = a;
a = b;
b = temp;
}
public void prnPermut(int[] list, int k, int m)
{
int i;
if (k == m)
{
for (i = 0; i <= m; i++)
Console.Write("{0}", list[i]);
Console.Write(" ");
}
else
for (i = k; i <= m; i++)
{
swapTwoNumber(ref list[k], ref list[i]);
prnPermut(list, k + 1, m);
swapTwoNumber(ref list[k], ref list[i]);
}
}
}
}
}
So, if I have 2 input values 1
and 2
, the above code will return the following results 12
and 21
.
I need it to return more results than that for example:
1
12
2
21
If I type in 3 values 1
and 2
and 3
, at the moment it returns:
123 132 213 231 321 312
But I need it to return all the double digit and single digit permutations at the same time.
Anyone know how I can go about doing that?
For example:
1 2 3 12 13 21 23 31 32 123 132 213 321 312
and so on, I may have missed out some combinations/permutations.
My end goal is to be able to do the same thing, but with strings, so if the input is one
and two
, the output would be:
one
onetwo
two
twoone
for three inputs such as:
one
and two
and three
the output would be:
one two three onetwo onethree twoone twothree threeone threetwo onetwothree onethreetwo twoonethree threetwoone threeonetwo
assuming i have not missed out any combinations.
Upvotes: 2
Views: 86
Reputation: 2203
This seems to do the job?
Test cases:
Input: 1,2 gives...
2 1 21 12
Input 1, 2, 3 gives...
3 2 32 23 1 31 13 21 12 321 312 231 213 123 132
Input "one", "two", "three" gives...
three two threetwo twothree one threeone onethree twoone onetwo threetwoone threeonetwo twothreeone twoonethree onetwothree onethreetwo
From this code:
class Program
{
public static void Main()
{
formPermut test = new formPermut();
test.prnPermutWithSubsets(new object[] { 1, 2 });
Console.WriteLine();
test.prnPermutWithSubsets(new object[] { 1, 2, 3 });
Console.WriteLine();
test.prnPermutWithSubsets(new string[] { "one", "two", "three" });
Console.WriteLine();
return;
}
class formPermut
{
private void swapTwoNumber(ref object a, ref object b)
{
object temp = a;
a = b;
b = temp;
}
public void prnPermutWithSubsets(object[] list)
{
for (int i = 0; i < Math.Pow(2, list.Length); i++)
{
Stack<object> combination = new Stack<object>();
for (int j = 0; j < list.Length; j++)
{
if ((i & (1 << (list.Length - j - 1))) != 0)
{
combination.Push(list[j]);
}
}
this.prnPermut(combination.ToArray(), 0, combination.Count() - 1);
}
}
public void prnPermut(object[] list, int k, int m)
{
int i;
if (k == m)
{
for (i = 0; i <= m; i++)
Console.Write("{0}", list[i]);
Console.Write(" ");
}
else
for (i = k; i <= m; i++)
{
swapTwoNumber(ref list[k], ref list[i]);
prnPermut(list, k + 1, m);
swapTwoNumber(ref list[k], ref list[i]);
}
}
}
}
Upvotes: 1