def number_to_roma(n)
#初始化数据
number = [1,4,5,9,10,40,50,90,100,400,500,900,1000]
roma = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"]
hash = number.zip(roma).inject({}) {|r,ele| r[ele.first] = ele.last;r}
#获得个、十、百、千位的数字,千位数有可能大于2位数,但目前这个用于10000以内的数字
temp = 3.step(0,-1).inject([]) do |r,ele|
r << n/(10**ele)
n %= 10**ele
r
end
#转置一下,方便下面的计算
temp.reverse!
#开始计算
result = ""
3.step(0,-1).each do |i|
#如果是千位数,直接用"M" * 千位数的值获得一个字符串
if i == 3
result << "M"*temp[i]
next
end
#temp[i]*(10**i)相当于把当前位数的值转换成对应的真实值,再判断该值是否在number中存在
#如果存在,则直接在hash中把该值对应的字符串取过来
#如果不存在,则先判断当前位数的值是否小于5
#如果小于5,则直接用当前位数对应的(100,10,1)对应的字符串(C,X,I) * 当前位数的值获得字符串
#如果大于5,则把(500,50,5)(D,L,V)对应的字符串 + (C,X,I) * 当前位数的值-5 获得字符串
if number.include?(temp[i]*(10**i))
result << hash[temp[i]*(10**i)]
else
temp[i].to_i < 5 ? result << hash[10**i]*temp[i] : result << (hash[5*(10**i)]+hash[10**i]*(temp[i]-5))
end
end
#返回结果
result
end
p number_to_roma(1661)
p number_to_roma(1990)
p number_to_roma(2008)
def roma_to_number(string)
#初始化数据
number = [1,4,5,9,10,40,50,90,100,400,500,900,1000]
roma = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"]
hash = roma.zip(number).inject({}) {|r,ele| r[ele.first] = ele.last; r}
#生成一个正则表达式
reg = /M|CM|D|CD|C|XC|L|XL|X|IX|V|IV|I/
#用scan方法来生成一个数组,该数组的元素由正则表达式中的字符组成
temp = string.scan(reg)
#去重后,计算每一个元素在原数组的个数,再将个数乘以该字符对应的数值。最后再相加得出最终值。
temp.uniq.inject(0) do |r,ele|
r += temp.count(ele)*hash[ele]
end
end
p roma_to_number("MDCLXI")
p roma_to_number("MCMXC")
p roma_to_number("MMVIII")
网友评论