11

可能重复：
列出字符串/整数的所有排列

例如，

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

基本上，想象一下计算二进制，但不是从 0 到 1，而是从 a 到 z。

我一直试图让这个工作无济于事，而且公式变得相当复杂。我不确定是否有更简单的方法来做到这一点。

编辑

我现在有这样的东西，但它并不完全存在，我不确定是否有更好的方法：

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;
            }
        }
    }
}
4

4 回答 4

33

对于可变数量的字母组合，您可以执行以下操作：

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);
于 2012-12-15T10:22:19.683 回答
14 
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);
}

_ _编辑_ _

对于一般解决方案，您可以在此处使用 CartesianProduct

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;
}
于 2012-12-15T09:59:52.190 回答
4

这是一个非常简单的解决方案：

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);
于 2012-12-15T10:08:03.480 回答
4

对于 3 个“数字”，您有 26^3 个计数。只需在三个循环中从“a”迭代到“z”。

于 2012-12-15T10:00:15.277 回答