12. Integer to Roman

Roman numerals are represented by seven different symbols: I,V,X,L,C,DandM.

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:

• Ican be placed before V

  (5) and X(10) to make 4 and 9.

• Xcan be placed before L(50) and C(100) to make 40 and 90.

• Ccan 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();
    }
}