这个问题有你转换成数字的话 - 300将转化为“三百”。我知道这个问题已经解决了堆栈溢出,但我正在寻求一些关于我的特定递归解决方案的建议。将数字转换的话红宝石 - 提高我的递归解决方案
我的解决方案适用于所有数字,除了相同的数字: 100,1000,百万
以上将返回为“百”,“千”,“万”。我明白为什么会这样。
300,3300,32000,35000,等等。任何数量的100%或1000或百万== 0将返回为 “3300零”。
我的唯一成功的想法是用第二种方法get_words
通过in_words
运行数量,如果数量是大于0之后,从解决方案中删除“零”,而刚刚回归“一” +号而不是通过in_words
运行如果数字是100/1000/1000000 ...
它可能是我的解决办法只能这样工作使用递归这里,但也许我失去了一些东西的局限性所致。任何人都可以提供一种方法来调整我的解决方案来解决这个问题?
NUMBER_DICTIONARY = {
0 => "zero",
1 => "one",
2 => "two",
3 => "three",
4 => "four",
5 => "five",
6 => "six",
7 => "seven",
8 => "eight",
9 => "nine",
10 => "ten",
11 => "eleven",
12 => "twelve",
13 => "thirteen",
14 => "fourteen",
15 => "fifteen",
16 => "sixteen",
17 => "seventeen",
18 => "eighteen",
19 => "nineteen",
20 => "twenty",
30 => "thirty",
40 => "forty",
50 => "fifty",
60 => "sixty",
70 => "seventy",
80 => "eighty",
90 => "ninety",
100 => "hundred",
1000 => "thousand",
1000000 => "million"
}
def get_words(number)
result = in_words(number)
return "one" + " " + NUMBER_DICTIONARY[number] if (number == 100 || number == 1000 || number == 1000000)
return result.split(" ").delete_if { |word| word == "zero"}.join(" ") if result.include?("zero") && number > 0
result
end
def in_words(number)
#base case
return NUMBER_DICTIONARY[number] if NUMBER_DICTIONARY.include?(number)
#recursion
return in_words(number/1000000) + " " + in_words(1000000) + " " + in_words(number % 1000000) if number > 1000000
return in_words(number/1000) + " " + in_words(1000) + " " + in_words(number % 1000) if number > 1000
return in_words(number/100) + " " + in_words(100) + " " + in_words(number % 100) if number > 100
return in_words((number/10) * 10) + " " + in_words(number % 10) if number > 20
end