def number_to_roma(n)
#初始化數(shù)據(jù)
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}
#獲得個(gè)、十、百、千位的數(shù)字删窒，千位數(shù)有可能大于2位數(shù)狂男，但目前這個(gè)用于10000以?xún)?nèi)的數(shù)字
temp = 3.step(0,-1).inject([]) do |r,ele|
r << n/(10**ele)
n %= 10**ele
r
end
#轉(zhuǎn)置一下崔泵，方便下面的計(jì)算
temp.reverse!
#開(kāi)始計(jì)算
result = ""
3.step(0,-1).each do |i|
#如果是千位數(shù)素标，直接用"M" * 千位數(shù)的值獲得一個(gè)字符串
if i == 3
result << "M"*temp[i]
next
end
#temp[i]*(10**i)相當(dāng)于把當(dāng)前位數(shù)的值轉(zhuǎn)換成對(duì)應(yīng)的真實(shí)值，再判斷該值是否在number中存在
#如果存在蒸痹，則直接在hash中把該值對(duì)應(yīng)的字符串取過(guò)來(lái)
#如果不存在，則先判斷當(dāng)前位數(shù)的值是否小于5
#如果小于5呛哟，則直接用當(dāng)前位數(shù)對(duì)應(yīng)的（100叠荠，10，1）對(duì)應(yīng)的字符串（C,X,I) * 當(dāng)前位數(shù)的值獲得字符串
#如果大于5竖共，則把（500蝙叛，50，5）（D,L,V)對(duì)應(yīng)的字符串 + （C,X,I) * 當(dāng)前位數(shù)的值-5 獲得字符串
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
#返回結(jié)果
result
end
p number_to_roma(1661)
p number_to_roma(1990)
p number_to_roma(2008)
def roma_to_number(string)
#初始化數(shù)據(jù)
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}
#生成一個(gè)正則表達(dá)式
reg = /M|CM|D|CD|C|XC|L|XL|X|IX|V|IV|I/
#用scan方法來(lái)生成一個(gè)數(shù)組公给，該數(shù)組的元素由正則表達(dá)式中的字符組成
temp = string.scan(reg)
#去重后借帘，計(jì)算每一個(gè)元素在原數(shù)組的個(gè)數(shù)，再將個(gè)數(shù)乘以該字符對(duì)應(yīng)的數(shù)值淌铐。最后再相加得出最終值肺然。
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")
