# 2022-06-20

## [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/)

### Description

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]`.

### Solution

#### Approach #0

```go
var (
    m = map[byte]int{
        'I': 1,
        'V': 5,
        'X': 10,
        'L': 50,
        'C': 100,
        'D': 500,
        'M': 1000,
    }
)

func romanToInt(s string) (ans int) {
    var last int
    for i := len(s) - 1; i >= 0; i-- {
        v := m[s[i]]
        if v >= last {
            ans += v
        } else {
            ans -= v
        }
        last = v
    }
    return
}
```