# LeetCode 13. Roman to Integer



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.&#x20;
* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.&#x20;
* `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
```

**Example 2:**

```
Input: s = "IV"
Output: 4
```

**Example 3:**

```
Input: s = "IX"
Output: 9
```

**Example 4:**

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

**Example 5:**

```
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:

[English Version in Youtube](https://youtu.be/53XFEgtot7g)

[中文版解答Youtube Link](https://youtu.be/LA3qRG3qS48)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1AN411d7bX/)

```
class Solution {
public:
    
    int romanToInt(string s) {
        map<char, int> table { 
            {'I', 1}, {'V', 5}, {'X', 10}, {'L', 50},
            {'C', 100}, {'D', 500}, {'M', 1000}
        };
        
        int result = 0;
        for (int i = 0;i < s.length();i++) {
            if (i < s.length()-1 && table[s[i]] < table[s[i+1]] ) {
                result -= table[s[i]];
            } else {
                result += table[s[i]];
            }
        }
        
        return result;
    }
    
};
```
