# Question

Formatted question description: https://leetcode.ca/all/12.html

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.

Example 1:

Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.


Example 2:

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


Example 3:

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


Constraints:

• 1 <= num <= 3999

# Algorithm

Because the range of input numbers (1-3999) is limited in the question, the question becomes much simpler.

I-1

V-5

X-10

L-50

C-100

D-500

M-1000


For example, the Roman numeral for the integer 1437 is MCDXXXVII. It is not difficult to find that the thousands, hundreds, tens, and ones digits are represented by Roman numerals. 1000-M, 400-CD, 30-XXX, 7-VII. So what we have to do is to use the quotient method to extract the numbers in each digit, and then express them separately:

100-C

200-CC

300-CCC

400-CD

500-D

600-DC

700-DCC

800-DCCC

900-CM


Can be divided into four categories, 100 to 300 category, 400 category, 500 to 800 category, 900 last category. The situation on digit is similar.

# Code

• 
public class Integer_to_Roman {
public class Solution {

/*
steps for Roman number:
1. 1 <= digit <= 3, append that digit number of Roman
2. digit = 4, append "I"+Roman
3. digit = 5,10, etc... append Roman
4. 6 <= digit <= 8, append Roman+"I"
5. digit = 9, append "I"+Roman
*/
public String intToRoman(int num) {

// @note: pre-calculate 900,400,90,40,9,4
String[] symbols = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};

// order: descending
int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};

StringBuilder result = new StringBuilder();

for (int i = 0; i < values.length; i++) {
while (num >= values[i]) {
result.append(symbols[i]);
num -= values[i];
}
}

return new String(result);

}
}
}

############

class Solution {
public String intToRoman(int num) {
int[] nums = new int[] {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String[] romans
= new String[] {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
StringBuilder sb = new StringBuilder();
for (int i = 0; i < nums.length; ++i) {
while (num >= nums[i]) {
num -= nums[i];
sb.append(romans[i]);
}
}
return sb.toString();
}
}

• // OJ: https://leetcode.com/problems/integer-to-roman/
// Time: O(1)
// Space: O(1)
class Solution {
string symbols = "MDCLXVI";
int values = {1000,500,100,50,10,5,1};
public:
string intToRoman(int num) {
string ans;
for (int i = 0; i < 7; i += 2) {
int d = num / values[i];
if (d == 0) continue;
if (d <= 3) {
while (d--) ans += symbols[i];
} else if (d == 4) {
ans += symbols[i];
ans += symbols[i - 1];
} else if (d == 5) {
ans += symbols[i - 1];
} else if (d <= 8) {
ans += symbols[i - 1];
for (int j = 5; j < d; ++j) ans += symbols[i];
} else {
ans += symbols[i];
ans += symbols[i - 2];
}
num %= values[i];
}
return ans;
}
};

• class Solution:
def intToRoman(self, num: int) -> str:
nums = [
(1000, 'M'),
(900, 'CM'),
(500, 'D'),
(400, 'CD'),
(100, 'C'),
(90, 'XC'),
(50, 'L'),
(40, 'XL'),
(10, 'X'),
(9, 'IX'),
(5, 'V'),
(4, 'IV'),
(1, 'I'),
]
res = []
for k, v in nums:
while num >= k:
num -= k
res.append(v)
return ''.join(res)

############

class Solution(object):
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
ans = ""
values = {"M": 1000, "D": 500, "C": 100, "L": 50, "X": 10, "V": 5, "I": 1}
literals = ["M", "D", "C", "L", "X", "V", "I"]
for idx in [0, 2, 4]:
k = num / values[literals[idx]]
re = (num % values[literals[idx]]) / values[literals[idx + 2]]
ans += k * literals[idx]
if re >= 9:
ans += literals[idx + 2] + literals[idx]
elif re >= 5:
ans += literals[idx + 1] + (re - 5) * literals[idx + 2]
elif re == 4:
ans += literals[idx + 2] + literals[idx + 1]
else:
ans += re * literals[idx + 2]
num %= values[literals[idx + 2]]
return ans


• func intToRoman(num int) string {
ans := ""
values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
romans := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
for i, value := range values {
for value <= num {
ans, num = ans+romans[i], num-value
}
}
return ans
}

• function intToRoman(num: number): string {
const nums: number[] = [
1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1,
];
const romans: string[] = [
'M',
'CM',
'D',
'CD',
'C',
'XC',
'L',
'XL',
'X',
'IX',
'V',
'IV',
'I',
];
let ans: string = '';
for (let i = 0; i < nums.length; ++i) {
while (num >= nums[i]) {
num -= nums[i];
ans += romans[i];
}
}
return ans;
}


• public class Solution {
public string IntToRoman(int num) {
List<string> cs = new List<string>{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
List<int> vs = new List<int>{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
StringBuilder ans = new StringBuilder();
for (int i = 0; i < cs.Count; i++) {
while (num >= vs[i]) {
ans.Append(cs[i]);
num -= vs[i];
}
}
return ans.ToString();
}
}