## Question

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 ones 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 a roman numeral, convert it to an integer.

**Example 1:**

```
Input: s = "III"
Output: 3
Explanation: III = 3.
```

**Example 2:**

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

**Example 3:**

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

**Constraints:**

`1 <= s.length <= 15`

`s`

contains only the characters`('I', 'V', 'X', 'L', 'C', 'D', 'M')`

.- It is
**guaranteed**that`s`

is a valid roman numeral in the range`[1, 3999]`

.

## Approach

- Put the mapping into a map for use ( better coed quality and easy to use);
- Check each char and its previous one and do the calculation
- Check if final char should be include according to the rules(if it's a 4/9/40.... With its previous code then this char should not be added)

## Code

class Solution { public int romanToInt(String s) { Map<Character, Integer> map = new HashMap<>(); map.put('I', 1); map.put('V', 5); map.put('X', 10); map.put('L', 50); map.put('C', 100); map.put('D', 500); map.put('M', 1000); int sum = 0; int i = 0; for (; i < s.length() - 1; i++) { char ch = s.charAt(i); char next = s.charAt(i + 1); if ((ch == 'I' && (next == 'X' || next == 'V')) || (ch == 'X' && (next == 'L' || next == 'C')) || (ch == 'C' && (next == 'D' || next == 'M'))) { sum += (map.get(next) - map.get(ch)); i++; } else { sum += map.get(ch); } } return i == s.length() ? sum : sum + map.get(s.charAt(s.length() - 1)); } }