算法:里程表/蛮力

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.

如果我传入一个从0到J的字符数组,并将长度设置为5,我想要的结果如00000,00001,00002 ...... 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

这不是相当的副本"递归,而不是多回路"但它相当接近.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返回一个IEnumerable<string>,但返回一个迭代器给了一个很好的接口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;
}