Question

我正在尝试创建一个函数，它将在传递索引时为我提供字母位置。它会像excel显示它的列一样。 A ... Z，AA，AB ....我写了下面的函数来得到Z的结果。它看起来像

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;
}

直到'Z'才能正常工作。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传递给这个函数时，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。

对此问题的任何输入都会很棒！

修改

这是Tordek建议的。但他的代码将失败，如52,78等数字。为此添加了解决方法，这是最终的工作代码。

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);
    }
}

Answer 1

看到这个问题：
Translate a column index into an Excel Column Name

或者这一个：
How to convert a column number (eg. 127) into an excel column (eg. AA)

虽然第一个链接在顶部有正确答案，而第二个链接有几个不正确。

Answer 2

任何递归函数都可以转换为等效的迭代函数。我发现首先递归思考总是很容易：

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;
}

即便如此，请听Joel。

Answer 3

递归是一种可能性 - 如果index > 26，您在此调用中处理index % 26并将其连接到index / 26上的递归调用。但是，迭代通常更快，并且不难安排这样的简单情况。在伪代码中：

string result = <convert `index % 26`>
while index > 26:
  index = index / 26
  result = <convert `index % 26`> + result
return result

等。

Answer 4

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

    if (index >= alphabetsCount)
    {
        result += GetColumnName(index-alphabetsCount)
    }
    return (string) (64 + index);
}

我的C＃是令人难以置信的。将其解释为伪代码 - 它几乎肯定不会编译，但可能会让你开始。

Answer 5

我不想在C＃中回答这个问题，但我会告诉你这在Haskell中是多么容易。

alphas :: [String]
alphas = [x ++ [c] | x <- ([]:alphas), c <- ['A'..'Z']]

Prelude> take 100 alphas
["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","AA","AB","AC","AD","AE","AF","AG","AH","AI","AJ","AK",
 "AL","AM","AN","AO","AP","AQ","AR","AS","AT","AU","AV","AW","AX","AY","AZ","BA",
 "BB","BC","BD","BE","BF","BG","BH","BI","BJ","BK","BL","BM","BN","BO","BP","BQ",
 "BR","BS","BT","BU","BV","BW","BX","BY","BZ","CA","CB","CC","CD","CE","CF","CG",
 "CH","CI","CJ","CK","CL","CM","CN","CO","CP","CQ","CR","CS","CT","CU","CV"]

增加字母表

5 个答案: