[LeetCode Road] Roman to Integer - Solution/C++

13. Roman to Integer

Question:

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:

Input: s = "\(III\)"
Output: 3

Input: s = "\(IV\)"
Output: 4

Input: s = "\(IX\)"
Output: 9

Input: s = "\(LVIII\)"
Output: 58
Explanation: \(L\) = 50, \(V\)= 5, \(III\) = 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].

Source code

Version 1

Idea:
At first, I thought these symbols can be built by using vector directly. After my careful thought, a map container is a easy way to store elements by a combination of a key value and a mapped value. Next step, I used a loop to check each characters, if symbol[s[i]] >= symbol[s[i + 1]] is true, add symbol[s[i]] up. On the contrary, if symbol[s[i]] < symbol[s[i + 1]] is true, subtract symbol[s[i]] from summation.

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

Version 2

Idea:
This is a way to reduce a computing time. Add symbol[s[i]] value first, then check a value of s[i - 1] element whether it is less than a value of s[i] or not. If it is true, the summation need to subtract twice value of symbol[s[i - 1]]. So, it implements the less conditional operator.

class Solution {
public:
    int romanToInt(string s) {
        map<char, int> symbol({{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100,}, {'D', 500}, {'M', 1000}});
        int sum = 0;
        for(int i = 0; i < s.length(); i++){
            sum += symbol[s[i]];
            if(i > 0 && symbol[s[i - 1]] < symbol[s[i]]){
                sum -= 2*symbol[s[i -1]];
            }
        }
        return sum;
    }
};