# Roman to Integer Leetcode Solution

In this post, we are going to solve the Roman to Integer Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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 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]`.

Now, lets see the leetcode solution of Roman to Integer Leetcode Solution.

### Roman to Integer Leetcode Solutionin Python

```class Solution(object):
def romanToInt(self, s):
"""
:type s: str
:rtype: int
"""
roman_dic = {
"I": 1,
"V": 5,
"X": 10,
"L": 50,
"C": 100,
"D": 500,
"M": 1000,
}

skip_next_n = False
int_sum = 0
for idx, n in enumerate(s):
if skip_next_n == True:
skip_next_n = False
continue
try:
if roman_dic[n] < roman_dic[s[idx + 1]]:
int_sum +=  roman_dic[s[idx + 1]] - roman_dic[n]
skip_next_n = True
continue
else:
int_sum += roman_dic[n]
except:
int_sum += roman_dic[n]

return int_sum```

### Roman to Integer Leetcode Solutionin CPP

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

}
};```

### Roman to Integer Leetcode Solution in Java

```public int romanToInt(String s) {
int ans = 0, num = 0;
for (int i = s.length()-1; i >= 0; i--) {
switch(s.charAt(i)) {
case 'I': num = 1; break;
case 'V': num = 5; break;
case 'X': num = 10; break;
case 'L': num = 50; break;
case 'C': num = 100; break;
case 'D': num = 500; break;
case 'M': num = 1000; break;
}
if (4 * num < ans) ans -= num;
else ans += num;
}
return ans;
}```

Note: This problem Roman to Integer is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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