递增字母 [英] Incrementing alphabets

问题描述

我正在尝试创建一个函数，该函数将在传递索引时为我提供字母位置。就像excel如何显示其列一样。 A ... Z，AA，AB ....我编写了以下函数，以将结果提高到Z。看起来

I am trying to create a function which will give me alphabet position when an index is passed. It will be same like how excel shows it's columns. A...Z, AA,AB.... I wrote the below function to get the results upto Z. It looks like

static string GetColumnName(int index)
{
    const int alphabetsCount = 26;
    if (index <= alphabetsCount)
    {
        int code = (index - 1) + (int)'A';
        return char.ConvertFromUtf32(code);
    }
    return string.Empty;
}

此功能有效，直到出现 Z。如果我通过1，则返回 A，如果我通过2，则返回 B，依此类推。但是，当我将27传递给该函数时，我无法弄清楚如何获得AA。我想我需要一个递归方法来找到它。

This works fine until 'Z'. It return 'A' if I pass 1 and return 'B' if I pass 2 and so on. But, I am not able to figure out how will I get AA when I pass 27 to this function. I guess I need a recursive method to find it.

对此问题的任何投入都会很棒！

Any inputs to this problem will be great!

编辑

这是Tordek建议的。但是他的代码将失败，例如52、78等。为此增加了解决方法，这是最终的工作代码。

This is suggested by Tordek. But his code will fail in numbers like 52, 78 etc. Added workaround for that and here is the final working code.

static string GetColumnName(int index)
{
    const int alphabetsCount = 26;

    if (index > alphabetsCount)
    {
        int mod = index % alphabetsCount;
        int columnIndex = index / alphabetsCount;

        // if mod is 0 (clearly divisible) we reached end of one combination. Something like AZ
        if (mod == 0)
        {
            // reducing column index as index / alphabetsCount will give the next value and we will miss one column.
            columnIndex -= 1;
            // passing 0 to the function will return character '@' which is invalid
            // mod should be the alphabets count. So it takes the last char in the alphabet.
            mod = alphabetsCount;
        }
        return GetColumnName(columnIndex) + GetColumnName(mod);
    }
    else
    {
        int code = (index - 1) + (int)'A';
        return char.ConvertFromUtf32(code);
    }
}

推荐答案

任何递归函数可以转换为等效的迭代函数。我发现总是很容易先进行递归思考：

Any recursive function can be converted into an equivalent iterative one. I find it always easy to think recursively first:

static string GetColumnName(int index)
{
    const int alphabetsCount = 26;

    if (index > alphabetsCount) {
        return GetColumnName(index / alphabetsCount) + GetColumnName(index % alphabetsCount);
    } else {
        int code = (index - 1) + (int)'A';
        return char.ConvertFromUtf32(code);
    }
}

可以简单地转换为：

static string GetColumnName(int index)
{
    const int alphabetsCount = 26;
    string result = string.Empty;

    while (index > 0) {
        result = char.ConvertFromUtf32(64 + (index % alphabetsCount)) + result;
        index /= alphabetsCount;
    }

    return result;
}

即使如此，也要听乔尔的话。

Even so, listen to Joel.

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