f#版本的各种方式

let rec getExcelColumnName x  = if x<26 then int 'A'+x|>char|>string else (x/26-1|>c)+ c(x%26)

对不起，最小化，正在开发一个更好的https://stackoverflow.com/a/4500043/57883版本

相反的方向:

// return values start at 0
let getIndexFromExcelColumnName (x:string) =
    let a = int 'A'
    let fPow len i =
        Math.Pow(26., len - 1 - i |> float)
        |> int

    let getValue len i c = 
        int c - a + 1 * fPow len i
    let f i = getValue x.Length i x.[i]
    [0 .. x.Length - 1]
    |> Seq.map f
    |> Seq.sum
    |> fun x -> x - 1

已经有30多个解决方案了，但这里是我的一行c#解决方案……

public string IntToExcelColumn(int i)
{
    return ((i<16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}

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

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;

打印稿

function lengthToExcelColumn(len: number): string {

    let dividend: number = len;
    let columnName: string = '';
    let modulo: number = 0;

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

这是我用python编写的解决方案

import math

num = 3500
row_number = str(math.ceil(num / 702))
letters = ''
num = num - 702 * math.floor(num / 702)
while num:
    mod = (num - 1) % 26
    letters += chr(mod + 65)
    num = (num - 1) // 26
result = row_number + ("".join(reversed(letters)))
print(result)