找出给定长度的所有可能单词的好方法是什么

Question

我正在尝试在C＃中创建一个生成以下输出字符串的算法：

AAAA
AAAB
AAAC
...and so on...
ZZZX
ZZZY
ZZZZ

实现这一目标的最佳方法是什么？

public static IEnumerable<string> GetWords()
{
    //Perform algorithm
    yield return word;
}

Answer 1

好吧，如果长度是常数4，那么这将处理它：

public static IEnumerable<String> GetWords()
{
    for (Char c1 = 'A'; c1 <= 'Z'; c1++)
    {
        for (Char c2 = 'A'; c2 <= 'Z'; c2++)
        {
            for (Char c3 = 'A'; c3 <= 'Z'; c3++)
            {
                for (Char c4 = 'A'; c4 <= 'Z'; c4++)
                {
                    yield return "" + c1 + c2 + c3 + c4;
                }
            }
        }
    }
}

如果长度是一个参数，这个递归解决方案将处理它：

public static IEnumerable<String> GetWords(Int32 length)
{
    if (length <= 0)
        yield break;

    for (Char c = 'A'; c <= 'Z'; c++)
    {
        if (length > 1)
        {
            foreach (String restWord in GetWords(length - 1))
                yield return c + restWord;
        }
        else
            yield return "" + c;
    }
}

Answer 2

总是有强制性的LINQ实现。最有可能是垃圾性能，但是从什么时候开始使用酷炫的新功能呢？

var letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();

var sequence = from one in letters
               from two in letters
               from three in letters
               from four in letters
               orderby one, two, three, four
               select new string(new[] { one, two, three, four });

'sequence'现在是一个包含AAAA到ZZZZ的IQueryable。

编辑：

好的，所以让我觉得应该可以使用LINQ制作一个可配置长度序列和可配置字母表。所以这就是。再一次，完全没有意义，但这让我烦恼。

public void Nonsense()
{
    var letters = new[]{"A","B","C","D","E","F",
                        "G","H","I","J","K","L",
                        "M","N","O","P","Q","R","S",
                        "T","U","V","W","X","Y","Z"};

    foreach (var val in Sequence(letters, 4))
        Console.WriteLine(val);
}

private IQueryable<string> Sequence(string[] alphabet, int size)
{
    // create the first level
    var sequence = alphabet.AsQueryable();

    // add each subsequent level
    for (var i = 1; i < size; i++)
        sequence = AddLevel(sequence, alphabet);

    return from value in sequence
           orderby value
           select value;
}

private IQueryable<string> AddLevel(IQueryable<string> current, string[] characters)
{
    return from one in current
           from character in characters
           select one + character;
}

对Sequence方法的调用产生与之前相同的AAAA到ZZZZ列表，但现在您可以更改使用的字典以及生成的字词的长度。

Answer 3

只是对Garry Shutler的讽刺，但我想要代码着色：

你真的不需要将它设为IQuaryable，也不需要排序，所以你可以删除第二种方法。一步转发就是使用Aggregate作为交叉产品，它最终会像这样：

IEnumerable<string> letters = new[]{
                "A","B","C","D","E","F",                       
                "G","H","I","J","K","L",
                "M","N","O","P","Q","R","S",           
                "T","U","V","W","X","Y","Z"};

var result = Enumerable.Range(0, 4)
                .Aggregate(letters, (curr, i) => curr.SelectMany(s => letters, (s, c) => s + c));

foreach (var val in result)
     Console.WriteLine(val);

Anders应该获得Linq的诺贝尔奖！

Answer 4

GNU Bash！

{a..z}{a..z}{a..z}{a..z}

Answer 5

受Garry Shutler的回答启发，我决定用T-SQL重新编写答案。

说“字母”是一个只有一个字段的表，MyChar，一个CHAR（1）。它有26行，每行都是字母。所以我们有（你可以在SQL Server上复制粘贴这些代码并按原样运行以查看它的运行情况）：

DECLARE @Letters TABLE (
    MyChar CHAR(1) PRIMARY KEY
)
DECLARE @N INT
SET @N=0
WHILE @N<26 BEGIN
    INSERT @Letters (MyChar) VALUES ( CHAR( @N + 65) )
    SET @N = @N + 1
END
-- SELECT * FROM @Letters ORDER BY 1

SELECT A.MyChar, B.MyChar, C.MyChar, D.MyChar
FROM @Letters A, Letters B, Letters C, Letters D
ORDER BY 1,2,3,4

优点是：它可以轻松扩展为使用大写/小写，或使用非英语拉丁字符（想想“Ñ”或cedille，eszets等），你仍然会得到一个有序集，只需要添加整理。另外，SQL Server在单核计算机上执行速度比LINQ略快，在多核（或多处理器）上执行可以并行执行，从而获得更大的提升。

不幸的是，它坚持了4个字母的具体情况。 lassevk的递归解决方案更为通用，尝试在T-SQL中进行通用解决方案必然意味着动态SQL具有其所有危险。

Answer 6

的Python！

（这只是一个黑客，不要太认真地对待我： - ）

# Convert a number to the base 26 using [A-Z] as the cyphers
def itoa26(n): 
   array = [] 
   while n: 
      lowestDigit = n % 26
      array.append(chr(lowestDigit + ord('A'))) 
      n /= 26 
   array.reverse() 
   return ''.join(array)

def generateSequences(nChars):
   for n in xrange(26**nChars):
      string = itoa26(n)
      yield 'A'*(nChars - len(string)) + string

for string in generateSequences(3):
   print string

Answer 7

的Haskell！

replicateM 4 ['A'..'Z']

红宝石！

('A'*4..'Z'*4).to_a

Answer 8

这是C＃中相同函数的递归版本：

using System;
using System.Collections.Generic;
using System.Text;
using System.IO;

namespace ConsoleApplication1Test
{
    class Program
    {
        static char[] my_func( char[] my_chars, int level)
        {
            if (level > 1)
                my_func(my_chars, level - 1);
            my_chars[(my_chars.Length - level)]++;
            if (my_chars[(my_chars.Length - level)] == ('Z' + 1))
            {
                my_chars[(my_chars.Length - level)] = 'A';
                return my_chars;
            }
            else
            {
                Console.Out.WriteLine(my_chars);
                return my_func(my_chars, level);
            }
        }
        static void Main(string[] args)
        {
            char[] text = { 'A', 'A', 'A', 'A' };
            my_func(text,text.Length);
            Console.ReadKey();
        }
    }
}

从AAAA打印到ZZZZ

Answer 9

更简单的Python！

def getWords(length=3):
    if length == 0: raise StopIteration
    for letter in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
        if length == 1: yield letter
        else:
            for partialWord in getWords(length-1):
                yield letter+partialWord

Answer 10

的JavaScript！

var chars = 4, abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ", top = 1, fact = [];
for (i = 0; i < chars; i++) { fact.unshift(top); top *= abc.length; }
for (i = 0; i < top; i++)
{
    for (j = 0; j < chars; j++) 
        document.write(abc[Math.floor(i/fact[j]) % abc.length]);
    document.write("<br \>\n");
}

Answer 11

使用自动Googles为每个单个字母组合的东西，然后查看是否有更多“.sz”或“.af”命中，然后“.com”命中五个第一个结果......;）

说真的，你正在寻找的东西可能是Tries（数据结构），虽然你仍然需要填充可能更难的东西......

Answer 12

一个非常简单但很棒的代码，可生成3个和4个英文字母的所有单词

#include <iostream>
using namespace std;
char alpha[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}
int main() {
int num;
cin >> num;
if (num == 3) { //all 3 letter words
    for (int i = 0; i <= 25; i++) {
        for (int o = 0; o <= 25; o++) {
            for (int p = 0; p <= 25; p++) {
                cout << alpha[i] << alpha[o] << alpha[p] << " ";
            }
        }
    }
}
else if (num == 4) { //all 4 letter words
    for (int i = 0; i <= 25; i++) {
        for (int o = 0; o <= 25; o++) {
            for (int p = 0; p <= 25; p++) {
                for (int q = 0; q <= 25; q++) {
                    cout << alpha[i] << alpha[o] << alpha[p] << alpha[q] << " ";
                }
            }
        }
    }
}
else {
    cout << "Not more than 4"; //it will take more than 2 hours for generating all 5 letter words
  }
}