Linq的组合发电机

Question

是否有可能创建一些Linq,生成一个包含一系列数字的所有可能组合的List？

如果输入"21",它将生成一个包含以下元素的列表:

list[0] = "21"
list[1] = "22"
list[2] = "11"
list[3] = "12"

(不按顺序)

我知道你可以使用范围来做以下事情:

List<char> letterRange = Enumerable.Range('a', 'z' - 'a' + 1).Select(i => (Char)i).ToList(); //97 - 122 + 1 = 26 letters/iterations

从az生成字母表.但我似乎无法转移这些知识来制作组合发生器

我已经能够用以下代码弄清楚它,但它似乎太笨重了,我相信它可以用几行完成.它确实感觉我做的是一个糟糕的解决方案.

想象一下,GetAllCombinations("4321")如果有帮助,我已经打过电话

public static String[] GetAllCombinations(String s)
{
    var combinations = new string[PossibleCombinations(s.Length)];

    int n = PossibleCombinations(s.Length - 1);

    for (int i = 0; i < s.Length; i++)
    {
        String sub;
        String[] subs;

        if (i == 0)
        {
            sub = s.Substring(1); //Get the first number
        }
        else if (i == s.Length - 1)
        {
            sub = s.Substring(0, s.Length - 1);
        }
        else
        {
            sub = s.Substring(0, i) + s.Substring(i + 1); 
        }

        subs = GetAllCombinations(sub);

        for (int j = 0; j < subs.Length; j++)
        {
            combinations[i * n + j] = s[i] + subs[j];
        }
    }

    return combinations;
}
public static int PossibleCombinations(int n) //Combination possibilities. e.g 1-2-3-4 have 24 different combinations
{
    int result = 1;

    for (int i = 1; i <= n; i++)
        result *= i;

    return result;
}

Answer 1

为了它的价值,尝试这样的事情:

public static IEnumerable<string> GetPermutations(string s)
{
    if (s.Length > 1)
        return from ch in s
               from permutation in GetPermutations(s.Remove(s.IndexOf(ch), 1))
               select string.Format("{0}{1}", ch, permutation);

    else
        return new string[] { s };
}

Answer 2

为了记录:Josh回答了通用的方式:

public static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> items) {
        if (items.Count() > 1) {
            return items.SelectMany(item => GetPermutations(items.Where(i => !i.Equals(item))),
                                   (item, permutation) => new[] { item }.Concat(permutation));
        } else {
            return new[] {items};
        }
    }

Answer 3

这是我使用Linq的Permutation和Combination功能

public static IEnumerable<TSource> Prepend<TSource>(this IEnumerable<TSource> source, TSource item)
{
    if (source == null)
        throw new ArgumentNullException("source");

    yield return item;

    foreach (var element in source)
        yield return element;
}

public static IEnumerable<IEnumerable<TSource>> Permutate<TSource>(this IEnumerable<TSource> source)
{
    if (source == null)
        throw new ArgumentNullException("source");

    var list = source.ToList();

    if (list.Count > 1)
        return from s in list
                from p in Permutate(list.Take(list.IndexOf(s)).Concat(list.Skip(list.IndexOf(s) + 1)))
                select p.Prepend(s);

    return new[] { list };
}

public static IEnumerable<IEnumerable<TSource>> Combinate<TSource>(this IEnumerable<TSource> source, int k)
{
    if (source == null)
        throw new ArgumentNullException("source");

    var list = source.ToList();
    if (k > list.Count)
        throw new ArgumentOutOfRangeException("k");

    if (k == 0)
        yield return Enumerable.Empty<TSource>();

    foreach (var l in list)
        foreach (var c in Combinate(list.Skip(list.Count - k - 2), k - 1))
            yield return c.Prepend(l);
}

对于DNA字母"A","C","G","T":

var dna = new[] {'A', 'C', 'G', 'T'};

foreach (var p in dna.Permutate())
    Console.WriteLine(String.Concat(p));

给

ACGT ACTG AGCT AGTC ATCG ATGC CAGT CATG CGAT CGTA CTAG CTGA GACT GATC GCAT GCTA GTAC GTCA TACG TAGC TCAG TCGA TGAC TGCA

和DNA字母表的组合(k = 2)

foreach (var c in dna.Combinate(2))
        Console.WriteLine(String.Concat(c));

是

AA AC AG AT CA CC CG CT GA GC GG GT TA TC TG TT