Question

我想用C＃风格的语言编写类似里程表的方法，但不只是使用0-9表示字符，而是使用任何字符集。它或多或少会像蛮力的应用程序一样。

如果我将 0 中的字符数组传递给 J ，并将长度设置为5，我想要 00000,00001,00002的结果...... HJJJJ，IJJJJJ，JJJJJ 。

这是基地，请帮我扩展：

protected void Main()
{
    char[] chars = new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
        'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J' };

    BruteForce(chars, 5);
}

private void BruteForce(char[] chars, int length)
{
    // for-loop (?) console-writing all possible combinations from 00000 to JJJJJ
    // (when passed in length is 5)
    // TODO: Implement code...
}

Answer 1

这不是完全 "recursion instead of multi-loops"的副本，但它非常接近。如果这对您没有帮助，我会写一个解决方案。

编辑：这是一个非递归的解决方案。递归的一个稍微难以从IEnumerable<string>返回，但是返回迭代器会给出一个很好的接口IMO：）

private static IEnumerable<string> GetAllMatches(char[] chars, int length)
{
    int[] indexes = new int[length];
    char[] current = new char[length];
    for (int i=0; i < length; i++)
    {
        current[i] = chars[0];
    }
    do
        {
            yield return new string(current);
        }
        while (Increment(indexes, current, chars));
}

private static bool Increment(int[] indexes, char[] current, char[] chars)
{
    int position = indexes.Length-1;

    while (position >= 0)
    {
        indexes[position]++;
        if (indexes[position] < chars.Length)
        {
             current[position] = chars[indexes[position]];
             return true;
        }
        indexes[position] = 0;
        current[position] = chars[0];
        position--;
    }
    return false;
}

Answer 2

这是我找到的解决方案之一。我喜欢它的紧凑性和分离性：

private static char[] characters =
    new char[] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J' };

// length: The length of the string created by bruteforce
public static void PerformBruteForce(int length) {
    int charactersLength = characters.Length;
    int[] odometer = new int[length];
    long size = (long)Math.Pow(charactersLength, length);

    for (int i = 0; i < size; i++) {
        WriteBruteForce(odometer, characters);
        int position = 0;
        do {
            odometer[position] += 1;
            odometer[position] %= charactersLength;
        } while (odometer[position++] == 0 && position < length);
    }
}

private static void WriteBruteForce(int[] odometer, char[] characters) {
    // Print backwards
    for (int i = odometer.Length - 1; i >= 0; i--) {
        Console.Write(characters[odometer[i]]);
    }
    Console.WriteLine();
}

Answer 3

谷歌的排列。

但是，如果您只是在处理“十六进制”范围，请执行以下操作：

for (int i = 0; i < (1 << 24); i++)
  string s = i.ToString("X6");

Answer 4

以下是我之前为此目的使用的一个类......顾名思义，它必须根据所提供的字符集中的字符数在不同的基数中计数。希望它有用......

public class BaseNCounter
{
    public char[] CharSet { get; set; }
    public int Power { get; set; }

    public BaseNCounter() { }

    public IEnumerable<string> Count() {
        long max = (long)Math.Pow((double)this.CharSet.Length, (double)this.Power);
        long[] counts = new long[this.Power];
        for(long i = 0; i < max; i++)
            yield return IncrementArray(ref counts, i);
    }

    public string IncrementArray(ref long[] counts, long count) {
        long temp = count;
        for (int i = this.Power - 1; i >= 0 ; i--) {
            long pow = (long)Math.Pow(this.CharSet.Length, i);
            counts[i] = temp / pow;
            temp = temp % pow;
        }

        StringBuilder sb = new StringBuilder();
        foreach (int c in counts) sb.Insert(0, this.CharSet[c]);
        return sb.ToString();
    }
}

在控制台应用程序中有几个使用场景。

class Program
{
    static void Main(string[] args)
    {
        BaseNCounter c = new BaseNCounter() { 
            CharSet = new char [] { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f' }, 
            Power = 2};

        foreach(string cc in c.Count())
            Console.Write("{0} ,", cc);
        Console.WriteLine("");

        BaseNCounter c2 = new BaseNCounter()
        {
            CharSet = new char[] { 'x', 'q', 'r', '9'},
            Power = 3
        };
        foreach (string cc in c2.Count())
            Console.Write("{0} ,", cc);
        Console.Read();
    }
}

Answer 5

我刚刚发表了一篇关于我在90年代做过的旧的（但非常棒的）暴力程序的重写，它可以从我的Gist中获得并且将完全按照你的要求行事要求：

https://gist.github.com/johanssonrobotics/11249060

祝你好运！

算法：里程表/蛮力

5 个答案: