P95 - 英文数字单词
English number words
官方模块：
Problems.P95核心函数：fullWords
题目描述
把非负整数逐位转换为英文单词，并用连字符连接。题目不是把 175 写成 “one hundred seventy-five”，而是 one-seven-five。
实现
方法一：算术拆分数字
1import Data.List (intercalate)
2
3fullWords :: Integral a => a -> String
4fullWords n
5 | n < 0 = error "fullWords: non-negative input required"
6 | n == 0 = "zero"
7 | otherwise = intercalate "-" (map digitWord (digits n))
8
9digits :: Integral a => a -> [Int]
10digits 0 = []
11digits n = digits (n `div` 10) ++ [fromIntegral (n `mod` 10)]
12
13digitWord 0 = "zero"
14digitWord 1 = "one"
15digitWord 2 = "two"
16digitWord 3 = "three"
17digitWord 4 = "four"
18digitWord 5 = "five"
19digitWord 6 = "six"
20digitWord 7 = "seven"
21digitWord 8 = "eight"
22digitWord 9 = "nine"
23digitWord _ = error "digitWord: not a digit"
方法二：复用十进制 show
1fullWordsViaShow :: Integral a => a -> String
2fullWordsViaShow n
3 | n < 0 = error "fullWords: non-negative input required"
4 | otherwise = intercalate "-"
5 [digitWord (fromEnum c - fromEnum '0') | c <- show (toInteger n)]
show 已经完成十进制位的提取，并且会为 0 产生字符串 "0"。随后只需把每个字符转换成数字下标并查单词。
方法对比
| 方法 | 数位来源 |
|---|---|
| 算术拆分 | 反复执行 div 和 mod |
show | 复用标准十进制格式化 |
测试
1>>> fullWords 175
2"one-seven-five"
3>>> fullWords 0
4"zero"
5>>> fullWords 1002
6"one-zero-zero-two"
7>>> fullWordsViaShow 1002
8"one-zero-zero-two"
digits 中的尾部 ++ 对位数很少的整数足够清楚；也可用累加器一次生成正序数字。