到目前为止，所有的解决方案都包含迭代或递归，这让我感到惊讶。

这是我的解，在常数时间内运行(没有循环)。此解决方案适用于所有可能的Excel列，并检查输入是否可以转换为Excel列。可能的列在[A, XFD]或[1,16384]范围内。(这取决于你的Excel版本)

private static string Turn(uint col)
{
    if (col < 1 || col > 16384) //Excel columns are one-based (one = 'A')
        throw new ArgumentException("col must be >= 1 and <= 16384");

    if (col <= 26) //one character
        return ((char)(col + 'A' - 1)).ToString();

    else if (col <= 702) //two characters
    {
        char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
        char secondChar = (char)(col % 26 + 'A' - 1);

        if (secondChar == '@') //Excel is one-based, but modulo operations are zero-based
            secondChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}", firstChar, secondChar);
    }

    else //three characters
    {
        char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
        char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
        char thirdChar = (char)(col % 26 + 'A' - 1);

        if (thirdChar == '@') //Excel is one-based, but modulo operations are zero-based
            thirdChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
    }
}

抱歉，这是Python而不是c#，但至少结果是正确的:

def excel_column_number_to_name(column_number):
    output = ""
    index = column_number-1
    while index >= 0:
        character = chr((index%26)+ord('A'))
        output = output + character
        index = index/26 - 1

    return output[::-1]


for i in xrange(1, 1024):
    print "%4d : %s" % (i, excel_column_number_to_name(i))

通过这些测试用例:

列号:494286 => ABCDZ 列号:27 => 列号:52 => AZ

我今天必须做这个工作，我的实现使用递归:

private static string GetColumnLetter(string colNumber)
{
    if (string.IsNullOrEmpty(colNumber))
    {
        throw new ArgumentNullException(colNumber);
    }

    string colName = String.Empty;

    try
    {
        var colNum = Convert.ToInt32(colNumber);
        var mod = colNum % 26;
        var div = Math.Floor((double)(colNum)/26);
        colName = ((div > 0) ? GetColumnLetter((div - 1).ToString()) : String.Empty) + Convert.ToChar(mod + 65);
    }
    finally
    {
        colName = colName == String.Empty ? "A" : colName;
    }

    return colName;
}

该方法将数字视为字符串，而以“0”开头的数字(A = 0)

这是一个javascript版本，根据格雷厄姆的代码

function (columnNumber) {
    var dividend = columnNumber;
    var columnName = "";
    var modulo;

    while (dividend > 0) {
        modulo = (dividend - 1) % 26;
        columnName = String.fromCharCode(65 + modulo) + columnName;
        dividend = parseInt((dividend - modulo) / 26);
    }

    return columnName;
};

int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
    int nChar = nCol % 26;
    nCol = (nCol - nChar) / 26;
    // You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
    sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;

更新:Peter的评论是正确的。这就是我在浏览器中编写代码的结果。:-)我的解决方案是不编译，它是最左边的字母，它是在反向顺序构建字符串-现在都固定了。

除了bug之外，该算法基本上是将一个数字从10进制转换为26进制。

更新2:Joel Coehoorn是对的-上面的代码将返回AB为27。如果它是一个以26为底的实数，AA就等于A, Z之后的下一个数字就是BA。

int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
    int nChar = nCol % 26;
    if (nChar == 0)
        nChar = 26;
    nCol = (nCol - nChar) / 26;
    sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
    sCol = sChars[nCol] + sCol;

如果你想以实用的方式引用单元格，那么如果你使用工作表的Cells方法，你会得到更可读的代码。它接受行和列索引，而不是传统的单元格引用。它与Offset方法非常相似。