如何获得所有可能的3个字母排列？
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，想象计算二进制，但不是从0到1，它从a到z。
我一直试图让这个工作几个小时现在无济于事，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。
谢谢阅读。
编辑：我现在有这样的东西，但它不是那里，我不确定是否有更好的方法：
private IEnumerable 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; } } } }
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);
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_ _
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> CartesianProduct(this IEnumerable> sequences) { // base case: IEnumerable> result = new[] { Enumerable.Empty () }; 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; }
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);
3“数字”你有26 ^ 3计数。 只需在三个循环中从’a’迭代到’z’。