Roman to Integer
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# LeetCode #13: Roman to Integer (C/C++).

easy

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.

SymbolValue
I1
V5
X10
L50
C100
D500
M1000

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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 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; } };