如何在HTML中将基数为10的数字转换为字母顺序列表

Question

我想将整数转换为字母等价,如HTML中的有序列表.

<ol type="a">

我尝试将基数为10的数字转换为带有az数字的基数26.
但这不是我想要的.

IN     WANT        GET      
-----------------------
1   =>  a       <=  a
2   =>  b       <=  b
3   =>  c       <=  c
4   =>  d       <=  d
5   =>  e       <=  e
6   =>  f       <=  f
7   =>  g       <=  g
8   =>  h       <=  h
9   =>  i       <=  i
10  =>  j       <=  j
11  =>  k       <=  k
12  =>  l       <=  l
13  =>  m       <=  m
14  =>  n       <=  n
15  =>  o       <=  o
16  =>  p       <=  p
17  =>  q       <=  q
18  =>  r       <=  r
19  =>  s       <=  s
20  =>  t       <=  t
21  =>  u       <=  u
22  =>  v       <=  v
23  =>  w       <=  w
24  =>  x       <=  x
25  =>  y       <=  y
26  =>  z       <=  az
27  =>  aa      <=  aa
28  =>  ab      <=  ab
29  =>  ac      <=  ac

---

private final static char[] digits = {
'0' , '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'
};

private static String numberToAlphaNumeric(long i, int radix) {

    char[] buf = new char[65];
    int charPos = 64;
    boolean negative = (i < 0);
    if (!negative) {
        i = -i;
    }
    while (i <= -radix) {
        buf[charPos--] = digits[(int)(-(i % radix))];
        i = i / radix;
    }
    buf[charPos] = digits[(int)(-i)];
    if (negative) { 
        buf[--charPos] = '-';
    }
    return new String(buf, charPos, (65 - charPos));
}

public static String numberToAlphaNumeric(long number) {
    ArrayList<String> list = new ArrayList<String>();
    for( int j = 0; list.size() != number; j++ ) {
        String alpha = numberToAlphaNumeric( j, digits.length );
        if(!alpha.contains( "0" )) {
            list.add( alpha );
        }
    }
    return list.get( list.size()-1 );
}

我的第二个想法:

如果我将一个新的前导符号扩展到数字并将我的数字转换为基数为27的数字,我在每个进位中都有新的符号,这是错误的,我可以过滤掉它们.

这是非常低效和丑陋的,但我没有更多的想法.常见的方法是什么？

Answer 1

这是基本算法.如果您需要更高效,请使用StringBuffer:

  public static String getAlpha(int num) {

    String result = "";
    while (num > 0) {
      num--; // 1 => a, not 0 => a
      int remainder = num % 26;
      char digit = (char) (remainder + 97);
      result = digit + result;
      num = (num - remainder) / 26;
    }

    return result;
  }

另一种方法是转换为base 26,然后为你得到的字符串中的每个字符添加97.