How to get all the possible 3 letter permutations? [duplicate]

问题

Possible Duplicate:
Listing all permutations of a string/integer

For example,

aaa .. aaz .. aba .. abz .. aca .. acz .. azz .. baa .. baz .. bba .. bbz .. zzz

Basically, imagine counting binary but instead of going from 0 to 1, it goes from a to z.

I have been trying to get this working for a few hours now to no avail and the formula is getting quite complex and I'm not sure if there's a simpler way to do it.

Thanks for reading.

Edit: I have something like this at the moment but it's not quite there and I'm not sure if there is a better way:

private IEnumerable<string> GetWordsOfLength(int length)
{
    char letterA = 'a', letterZ = 'z';

    StringBuilder currentLetters = new StringBuilder(new string(letterA, length));
    StringBuilder endingLetters = new StringBuilder(new string(letterZ, length));

    int currentIndex = length - 1;

    while (currentLetters.ToString() != endingLetters.ToString())
    {
        yield return currentLetters.ToString();

        for (int i = length - 1; i > 0; i--)
        {
            if (currentLetters[i] == letterZ)
            {
                for (int j = i; j < length; j++)
                {
                    currentLetters[j] = letterA;
                }

                if (currentLetters[i - 1] != letterZ)
                {
                    currentLetters[i - 1]++;
                }
            }
            else
            {
                currentLetters[i]++;

                break;
            }
        }
    }
}

回答1:

For a variable amount of letter combinations, you can do the following:

var alphabet = "abcdefghijklmnopqrstuvwxyz";
var q = alphabet.Select(x => x.ToString());
int size = 4;
for (int i = 0; i < size - 1; i++)
    q = q.SelectMany(x => alphabet, (x, y) => x + y);

foreach (var item in q)
    Console.WriteLine(item);

回答2:

var alphabet = "abcdefghijklmnopqrstuvwxyz";
//or var alphabet = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (char)i);

var query = from a in alphabet
            from b in alphabet
            from c in alphabet
            select "" + a + b + c;

foreach (var item in query)
{
    Console.WriteLine(item);
}

__EDIT__

For a general solution, you can use the CartesianProduct here

int N = 4;
var result = Enumerable.Range(0, N).Select(_ => alphabet).CartesianProduct();
foreach (var item in result)
{
    Console.WriteLine(String.Join("",item));
}

// Eric Lippert’s Blog
// Computing a Cartesian Product with LINQ
// http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences)
{
    // base case: 
    IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
    foreach (var sequence in sequences)
    {
        var s = sequence; // don't close over the loop variable 
        // recursive case: use SelectMany to build the new product out of the old one 
        result =
            from seq in result
            from item in s
            select seq.Concat(new[] { item });
    }
    return result;
}

回答3:

Here's a very simple solution:

for(char first = 'a'; first <= (int)'z'; first++)
    for(char second = 'a'; second <= (int)'z'; second++)
        for(char third = 'a'; third <= (int)'z'; third++)
            Console.WriteLine(first.ToString() + second + third);

回答4:

You have 26^3 counts for 3 "digits". Just iterate from 'a' to 'z' in three loops.

来源：https://stackoverflow.com/questions/13891200/how-to-get-all-the-possible-3-letter-permutations

标签

algorithm

alphabetical