## 算法33--Integer to Roman

Roman numerals are represented by seven different symbols: `I``V``X``L``C``D` and `M`.

``````Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000``````

For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:

• `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
• `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
• `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

 数字 g个 s十 b百 q千 1--3 I--III X---XXX C---CCC M--MMM 4 IX XL CD 5 V L D 6--8 VI--VIII LX---LXXX DC---DCCC 9 IX XC CM 0 k空
``````class Solution:
def intToRoman(self, num):
s10 = ['I','X','C','M']
s5 = ['V','L','D']
tmp = 1000
t = 0
r = ''
index = 3
while num>0:
t = int(num/tmp)
#print(t, tmp, index)
if t>=1 and t<=3:
r = r + s10[index]*t
elif t==4:
r = r + s10[index] + s5[index]
elif t==5:
r = r + s5[index]
elif t>=6 and t<=8:
r = r + s5[index] + s10[index]*(t-5)
elif t==9:
r = r + s10[index] +s10[index+1]
num = num % tmp
tmp = tmp/10
index -= 1
return r``````

s10保存十进制的1,10,100,1000取值

s5保存5位置的5,50,500情况

r保存最终结果

``````public static String intToRoman(int num) {
String M[] = {"", "M", "MM", "MMM"};
String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
}``````