生成单词所有变体的算法

Question

我想通过下面的例子来解释我的问题。

假设单词：abc a 有变体：ä、à
b 没有变体。
c 有变体：ç

所以可能的词是：

abc
äbc
àbc
abç
äbç
àbç

现在我正在寻找一种算法，它可以打印带有任意字母变体的 abritray 单词的所有单词变体。

Answer 1

我建议您递归地解决这个问题。下面是一些 Java 代码供您入门：

static Map<Character, char[]> variants = new HashMap<Character, char[]>() {{
    put('a', new char[] {'ä', 'à'});
    put('b', new char[] {        });
    put('c', new char[] { 'ç'    });
}}; 

public static Set<String> variation(String str) {

    Set<String> result = new HashSet<String>();

    if (str.isEmpty()) {
        result.add("");
        return result;
    }

    char c = str.charAt(0);
    for (String tailVariant : variation(str.substring(1))) {
        result.add(c + tailVariant);
        for (char variant : variants.get(c))
            result.add(variant + tailVariant);
    }

    return result;
}

测试：

public static void main(String[] args) {
    for (String str : variation("abc"))
        System.out.println(str);
}

输出：

abc
àbç
äbc
àbc
äbç
abç

Answer 2

在 Python 中快速破解的解决方案：

def word_variants(variants):
  print_variants("", 1, variants);

def print_variants(word, i, variants):
  if i > len(variants):
    print word
  else:
    for variant in variants[i]:
      print_variants(word + variant, i + 1, variants)

variants = dict()
variants[1] = ['a0', 'a1', 'a2']
variants[2] = ['b0']
variants[3] = ['c0', 'c1']

word_variants(variants)

Answer 3

公共部分：

string[] letterEquiv = { "aäà", "b", "cç", "d", "eèé" };

// Here we make a dictionary where the key is the "base" letter and the value is an array of alternatives
var lookup = letterEquiv
    .Select(p => p.ToCharArray())
    .SelectMany(p => p, (p, q) => new { key = q, values = p }).ToDictionary(p => p.key, p => p.values);

用 C# 编写的递归变体。

List<string> resultsRecursive = new List<string>();

// I'm using an anonymous method that "closes" around resultsRecursive and lookup. You could make it a standard method that accepts as a parameter the two.
// Recursive anonymous methods must be declared in this way in C#. Nothing to see.
Action<string, int, char[]> recursive = null;
recursive = (str, ix, str2) =>
{
    // In the first loop str2 is null, so we create the place where the string will be built.
    if (str2 == null)
    {
        str2 = new char[str.Length];
    }

    // The possible variations for the current character 
    var equivs = lookup[str[ix]];

    // For each variation
    foreach (var eq in equivs)
    {
        // We save the current variation for the current character
        str2[ix] = eq;

        // If we haven't reached the end of the string
        if (ix < str.Length - 1)
        {
            // We recurse, increasing the index
            recursive(str, ix + 1, str2);
        }
        else
        {
            // We save the string
            resultsRecursive.Add(new string(str2));
        }
    }
};

// We launch our function
recursive("abcdeabcde", 0, null);

// The results are in resultsRecursive

非递归版本

List<string> resultsNonRecursive = new List<string>();

// I'm using an anonymous method that "closes" around resultsNonRecursive and lookup. You could make it a standard method that accepts as a parameter the two.
Action<string> nonRecursive = (str) =>
{
    // We will have two arrays, of the same length of the string. One will contain
    // the possible variations for that letter, the other will contain the "current"
    // "chosen" variation of that letter
    char[][] equivs = new char[str.Length][];
    int[] ixes = new int[str.Length];

    for (int i = 0; i < ixes.Length; i++)
    {
        // We start with index -1 so that the first increase will bring it to 0
        equivs[i] = lookup[str[i]];
        ixes[i] = -1;
    }

    // The current "workin" index of the original string
    int ix = 0;

    // The place where the string will be built.
    char[] str2 = new char[str.Length];

    // The loop will break when we will have to increment the letter with index -1
    while (ix >= 0)
    {
        // We select the next possible variation for the current character
        ixes[ix]++;

        // If we have exausted the possible variations of the current character
        if (ixes[ix] == equivs[ix].Length)
        {
            // Reset the current character to -1
            ixes[ix] = -1;

            // And loop back to the previous character
            ix--;

            continue;
        }

        // We save the current variation for the current character
        str2[ix] = equivs[ix][ixes[ix]];

        // If we are setting the last character of the string, then the string
        // is complete
        if (ix == str.Length - 1)
        {
            // And we save it
            resultsNonRecursive.Add(new string(str2));
        }
        else
        {
            // Otherwise we have to do everything for the next character 
            ix++;
        }
    }
};

// We launch our function
nonRecursive("abcdeabcde");

// The results are in resultsNonRecursive

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