# 12. Integer to Roman (Medium)

https://leetcode.com/problems/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: L = 50, V = 5, III = 3.
```

Example 5:

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

## Solutions

``````class Solution {

public String intToRoman(int num) {

// Build the map between the symbols and corresponding values
int[] values = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000};
String[] symbols = {"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C", "CD", "D", "CM", "M"};

StringBuilder sb = new StringBuilder();
for (int i = symbols.length - 1; i >= 0; i--) {

// Bigger divisor comes first.
int tmp = num / values[i];
num -= values[i] * tmp;

// if tmp = 0, can not be divided by current divisor, skip and move to smaller divisor
for (int j = 0; j < tmp; j++) {
sb.append(symbols[i]);
}
}

return sb.toString();
}
}
``````