Solution: Integer to Roman

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Leetcode Problem #12 (Medium): Integer to Roman

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Description:

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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, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 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.

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Examples:

Example 1:
Input:	num = 3
Output:	"III"

Example 2:
Input:	num = 4
Output:	"IV"

Example 3:
Input:	num = 9
Output:	"IX"

Example 4:
Input:	num = 58
Output:	"LVIII"
Explanation:	L = 50, V = 5, III = 3.

Example 5:
Input:	num = 1994
Output:	"MCMXCIV"
Explanation:	M = 1000, CM = 900, XC = 90 and IV = 4.

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Constraints:

• 1 <= num <= 3999

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Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

Just like Roman to Integer, this problem is most easily solved using a lookup table for the conversion between digit and numeral. In this case, we can easily deal with the values in descending order and insert the appropriate numeral (or numerals) as many times as we can while reducing the our target number (N) by the same amount.

Once N runs out, we can return ans.

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Implementation:

Java's StringBuilder can take care of repeated string concatenations without some of the overhead of making string copies.

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Javascript Code:

(Jump to: Problem Description || Solution Idea)

const val = [1000,900,500,400,100,90,50,40,10,9,5,4,1]
const rom = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"]

var intToRoman = function(N) {
    let ans = ""
    for (let i = 0; N; i++)
        while (N >= val[i]) ans += rom[i], N -= val[i]
    return ans
};

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Python Code:

(Jump to: Problem Description || Solution Idea)

val = [1000,900,500,400,100,90,50,40,10,9,5,4,1]
rom = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"]

class Solution:
    def intToRoman(self, N: int) -> str:
        ans = ""
        for i in range(13):
            while N >= val[i]:
                ans += rom[i]
                N -= val[i]
        return ans

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Java Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
    final static int[] val = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    final static String[] rom = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};

    public String intToRoman(int N) {
        StringBuilder ans = new StringBuilder();
        for (int i = 0; N > 0; i++)
            while (N >= val[i]) {
                ans.append(rom[i]);
                N -= val[i];
            }
        return ans.toString();
    }
}

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C++ Code:

(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    const int val[13] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    const string rom[13] = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};

    string intToRoman(int N) {
        string ans = "";
        for (int i = 0; N; i++)
            while (N >= val[i]) ans += rom[i], N -= val[i];
        return ans;
    }
};

Comments

[InterestedCoder]

What is the time and space complexity of this algorithm? Thanks

[張旭]

O(13) ?

LOL ....