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.
• `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
```

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]`.

Idea:

• Accumulate the value of each symbol.
• If the current symbol is greater than the previous one, substract twice of the previous value. e.g. IX, 1 + 10 – 2 * 1 = 9

Solution:

• Time complexity: O(n)
• Space complexity: O(1)
``````/**
* @param {string} s
* @return {number}
*/
var romanToInt = function(s) {
let conversion = {"I": 1, "V":5,"X":10,"L":50,"C":100,"D":500,"M":1000};
let total = 0;

for (var i = 0; i < s.length; i++) {
// first symbol
total += conversion[s[i]];
// if current symbol is bigger than previous symbol
// subtract 2 times of the previous value
if (i > 0 && conversion[s[i]] > conversion[s[i - 1]])
total -= 2 * conversion[s[i - 1]];
}