C# 递增字母表

我正在尝试创建一个函数，当传递索引时，该函数将为我提供字母表位置。它将与excel显示其列的方式相同。A…Z，AA，AB。。。。我编写了下面的函数来得到Z的结果

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;
}

---

递归是一种可能性——如果

index>26

，则在此调用中处理

index%26

，并将其连接到

index/26

上的递归调用。然而，对于像这样的简单情况，迭代通常更快，也不难安排。在伪代码中：

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

字符串结果=
当指数>26时：
指数=指数/26
结果=+结果
返回结果

等等

static string GetColumnName(int index) { const int alphabetsCount = 26; string result = ''; if (index >= alphabetsCount) { result += GetColumnName(index-alphabetsCount) } return (string) (64 + index); } 静态字符串GetColumnName（int索引） { 常量int字母计数=26； 字符串结果=“”； 如果（索引>=字母计数） { 结果+=GetColumnName（索引字母计数） } 返回（字符串）（64+索引）； } 我的C#又可怕又生锈。将其解释为伪代码-它几乎肯定不会编译，但可能会让您开始使用。

请参见以下问题：

或者这个：

---

尽管第一个链接在顶部有一个正确答案，第二个链接有几个不正确的答案。

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的：

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，听Joel说。

我不想用C#来回答这个问题，但我要告诉你在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"]

alphas:：[String]
alphas=[x++[c]|x伪代码没有问题，但重要的是要知道，如果要在循环中附加到字符串，可能应该使用StringBuilder类来保存对象分配：这是错误的，因为它不是一个基26系统。如果a为0，a将等于AA（0==00）。如果A是1，从Z到AA就像是从9到11。Paul，StringBuilder太过分了。32位int是2^32，26^7溢出，因此最大长度为7个字母。7次迭代几乎不费力。64位的长度只有14个字母。：）太好了。我的数学太生疏了。因此无法计算模运算。谢谢你的代码。
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"]