In this post, we are going to solve the 13. Roman to Integer problem of Leetcode. This problem 13. Roman to Integer is a Leetcode easy level problem. Let’s see code, 13. Roman to Integer.
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 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:
Ican be placed beforeV(5) andX(10) to make 4 and 9.Xcan be placed beforeL(50) andC(100) to make 40 and 90.Ccan be placed beforeD(500) andM(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 <= 15scontains 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, let’s see the code of 13. Roman to Integer – Leetcode Solution.
Roman to Integer – Leetcode Solution
13. Roman to Integer – Solution in Java
class Solution {
public static int romanToInt(String s) {
if (s == null || s.length() == 0)
return -1;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
map.put('I', 1);
map.put('V', 5);
map.put('X', 10);
map.put('L', 50);
map.put('C', 100);
map.put('D', 500);
map.put('M', 1000);
int len = s.length(), result = map.get(s.charAt(len - 1));
for (int i = len - 2; i >= 0; i--) {
if (map.get(s.charAt(i)) >= map.get(s.charAt(i + 1)))
result += map.get(s.charAt(i));
else
result -= map.get(s.charAt(i));
}
return result;
}
}
13. Roman to Integer – Solution in C++
class Solution {
public:
int romanToInt(string s)
{
unordered_map<char, int> T = { { 'I' , 1 },
{ 'V' , 5 },
{ 'X' , 10 },
{ 'L' , 50 },
{ 'C' , 100 },
{ 'D' , 500 },
{ 'M' , 1000 } };
int sum = T[s.back()];
for (int i = s.length() - 2; i >= 0; --i)
{
if (T[s[i]] < T[s[i + 1]])
{
sum -= T[s[i]];
}
else
{
sum += T[s[i]];
}
}
return sum;
}
};
13. Roman to Integer – Solution in Python
class Solution:
def romanToInt(self, s: str) -> int:
translations = {
"I": 1,
"V": 5,
"X": 10,
"L": 50,
"C": 100,
"D": 500,
"M": 1000
}
number = 0
s = s.replace("IV", "IIII").replace("IX", "VIIII")
s = s.replace("XL", "XXXX").replace("XC", "LXXXX")
s = s.replace("CD", "CCCC").replace("CM", "DCCCC")
for char in s:
number += translations[char]
return number
Note: This problem 13. Roman to Integer is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.