Leetcode - Roman to Integer

Problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

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

I used 'object' to group key and value. According to the constraints, if the next key is higher than the current key, it's required to reduce the current value from the next value (IV: 5-1, CM: 1000-100 ...).

1. Set map object and result.
2. Comparemap[s[i]] to map[s[i+1]] and if it's lower, deduct map[s[i]] from map[s[i+1]]. Otherwise add the current value(map[s[i]]) to the result.
3. Return the result

code

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function(s) {
    let map = {"I":1, "V":5, "X":10, "L":50, "C":100, "D":500, "M":1000}
    let result = 0;
    for(let i=0; i<s.length; i++) {
        let cur = map[s[i]]
        let next = map[s[i+1]]
        if(cur < next) {
            result += next-cur
            i++
        } else {
            result += cur
        }
    }
    return result
};