Incrementing字母

我想创建一个函数，当索引通过时，它会给我字母表位置。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Incrementing字母

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; 
}

这工作正常，直到'Z'。如果我传递1，则返回'A'，如果传递2，则返回'B'等。但是，我无法弄清楚当我将27传递给这个函数时如何获得AA。我想我需要一个递归的方法来找到它。

对这个问题的任何投入将是伟大的！

编辑

这是由Tordek建议。但他的代码会失败，像52,78等数字。增加了解决方法，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); 
    } 
}

来源

2009-05-29 Navaneeth K N

任何递归函数都可以转换为等价的迭代函数。我发现它总是容易的递归首先想到：

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); 
    } 
}

这可以简单转化为：

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，听乔尔。

来源

2009-05-29 02:33:24 Tordek

递归是一种可能性 - 如果index > 26，你在此调用处理index % 26，并串连到index/26递归调用。但是，迭代通常更加快速，并且不难安排诸如此类的简单情况。伪代码：

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

或类似的。

来源

2009-05-29 02:20:48

这没有什么错你的伪代码的，但是要知道，你应该使用StringBuilder类，如果你将要追加到一个字符串在一个循环，以节省对象分配是很重要的：HTTP：/ /msdn.microsoft.com/en-us/library/system.text.stringbuilder(loband).aspx – 2009-05-29 02:30:06

这是错误的，因为它不是基本的26系统。如果A为0，则A等于AA（0 == 00）。如果A是1，从Z到AA将会像从9到11. – 2009-05-29 02:31:28

 
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＃是可怕和坚定的。将其解释为伪代码 - 它几乎肯定不会编译，但可能会让您开始。

来源

2009-05-29 02:24:39

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)

虽然第一环节都有一个正确的答案就在顶部和第二有几个是不正确的。

来源

2009-05-29 02:26:15

-1

我不想在C＃中回答这个问题，但我会告诉你在Haskell中这是多么容易。

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"]

来源

2009-05-29 03:34:28

Incrementing字母

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