## Monday, July 13, 2015

### LeetCode OJ - Roman to Integer

Problem:

Solution:

With the previous problem solved, this one is easy. The key is to segment the roman string. The string is mostive additive, but in case a larger letter comes after, then it becomes subtraction. Just special case those subtraction case and we are good.

The great thing about this problem is that with the previous one, we can easily brute force testing!

Code:

#include "stdafx.h"

// https://leetcode.com/problems/roman-to-integer/

#include "LEET_ROMAN_TO_INTEGER.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_ROMAN_TO_INTEGER
{
class Solution
{
public:
int romanToInt(string s)
{
int value = 0;
for (unsigned int i = 0; i < s.size(); i++)
{
char currentCharacter = s[i];
char nextCharacter = '\0';
if (i != s.size() - 1)
{
nextCharacter = s[i + 1];
}
if (currentCharacter == 'M')
{
value += 1000;
}
else if (currentCharacter == 'D')
{
value += 500;
}
else if (currentCharacter == 'C')
{
if (nextCharacter == 'M')
{
i++;
value += 900;
}
else if (nextCharacter == 'D')
{
i++;
value += 400;
}
else
{
value += 100;
}
}
else if (currentCharacter == 'L')
{
value += 50;
}
else if (currentCharacter == 'X')
{
if (nextCharacter == 'C')
{
i++;
value += 90;
}
else if (nextCharacter == 'L')
{
i++;
value += 40;
}
else
{
value += 10;
}
}
else if (currentCharacter == 'V')
{
value += 5;
}
else if (currentCharacter == 'I')
{
if (nextCharacter == 'X')
{
i++;
value += 9;
}
else if (nextCharacter == 'V')
{
i++;
value += 4;
}
else
{
value += 1;
}
}
}
return value;
}
};
};

#include "LEET_INTEGER_TO_ROMAN.h"

using namespace _LEET_ROMAN_TO_INTEGER;
using namespace _LEET_INTEGER_TO_ROMAN;

int LEET_ROMAN_TO_INTEGER()
{
Solution s;
for (int i = 1; i < 4000; i++)
{
cout << (s.romanToInt(intToRoman(i)) == i) << endl;
}

return 0;
}