我想将整数转换为字母等价,如
HTML中的有序列表.
- <ol type="a">
我试图将基数为10的数字转换为带有-z数字的基数26.
但那不是我想要的.
- 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：
如果我将新的前导符号扩展为数字并将我的数字转换为基数为27的数字,
我在每次携带中都有新的符号,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？
解决方法
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;
- }
