**source:**https://leetcode.com/problems/roman-to-integer/**C/C++**

**Solution to LeetCode**problem

**13**.

**Roman to Integer**.

## Problem

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.

## Examples

**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

- In a hash table we store the roman letter with its value.
- We initialize a number with
`0`

. - We read every charcter from the roman number.
- If the current character has a lower value than the next character, then we substract the value of the current character from the value of the next character (i.e.
`XL`

–>`L - X`

–>`50 - 10`

). - If not, then we add the value of the next
`n`

characters as long as they are the same (i.e.`XX`

–>`X + X`

–>`10 + 10`

) - We add the obtained value to the initial number/result.

- If the current character has a lower value than the next character, then we substract the value of the current character from the value of the next character (i.e.

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class Solution {
public:
int romanToInt(string s) {
unordered_map<char, int> roman;
roman['I'] = 1;
roman['V'] = 5;
roman['X'] = 10;
roman['L'] = 50;
roman['C'] = 100;
roman['D'] = 500;
roman['M'] = 1000;
int i = 0;
int r = 0;
int c = 0;
while (i<s.size()) {
if (i < s.size() - 1 && roman[s[i]] < roman[s[i+1]]) {
r += (roman[s[i+1]] - roman[s[i]]);
i+=2;
continue;
}
c = roman[s[i]];
i ++;
while (i < s.size() &&roman[s[i]] == roman[s[i-1]]) {
c += roman[s[i]];
i++;
}
r += c;
}
return r;
}
};