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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can 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 <= 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, let’s see the leetcode solution of Roman to Integer Leetcode Solution.
Roman to Integer Leetcode Solution in 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 Solution in 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.