## 12. 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.

Example 1:

``````Input: 3

Output: "III"
``````

Example 2:

``````Input: 4

Output: "IV"
``````

Example 3:

``````Input: 9

Output: "IX"
``````

Example 4:

``````Input: 58

Output: "LVIII"

Explanation: C = 100, L = 50, XXX = 30 and III = 3.
``````

Example 5:

``````Input: 1994

Output: "MCMXCIV"

Explanation:
M = 1000, CM = 900, XC = 90 and IV = 4.
``````

Thoughts:

1. Most straightforward way is to enumerate all 1th, 10th, 100 th, 1000 th units expressions
2. Iterative solution:

Code: Enumerate

``````class Solution(object):
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
# Input is guaranteed to be within the range from 1 to 3999
M = ['', 'M', 'MM', 'MMM'] # 1000s
C = ['', 'C', 'CC', 'CCC', 'CD', 'D', 'DC','DCC','DCCC','CM'] # 100s
X = ['', 'X', 'XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC'] # 10s
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]
``````

Code: Enumerate

``````class Solution {
public String intToRoman(int num) {
String[] romanPieces={"","I","II","III","IV","V","VI","VII","VIII","IX",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","M","MM","MMM","MMMM"};
return romanPieces[num/1000+30]+romanPieces[(num/100)%10+20]
+romanPieces[(num/10)%10+10]+romanPieces[num%10];
}
}
``````

Code: Enumerate

``````class Solution(object):
def intToRoman(self, num):
values = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ]
numerals = [ "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" ]
res = ""
for i, v in enumerate(values):
res += (num/v) * numerals[i]
num %= v
return res
``````

Code: Iterative:

``````class Solution {
public String intToRoman(int num) {
int[] weights={1000,900,500,400,100,90,50,40,10,9,5,4,1};
String[] tokens={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
StringBuilder sb=new StringBuilder("");
int start=0;
while(num>0){
for(int i=start;i<13;i++){
if(num>=weights[i]){
num-=weights[i];
sb.append(tokens[i]);
break;
}
start=i+1; // skip those impossible check, make it faster
}
}
return sb.toString();
}
}
``````