source: https://leetcode.com/problems/roman-to-integer/
C/C++ Solution to LeetCode problem 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 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.
Examples
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]
.
Solution
- In a hash table we store the roman letter with its value.
- We initialize a number with
0
. - We read every charcter from the roman number.
- If the current character has a lower value than the next character, then we substract the value of the current character from the value of the next character (i.e.
XL
–>L - X
–>50 - 10
). - If not, then we add the value of the next
n
characters as long as they are the same (i.e.XX
–>X + X
–>10 + 10
) - We add the obtained value to the initial number/result.
- If the current character has a lower value than the next character, then we substract the value of the current character from the value of the next character (i.e.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
class Solution {
public:
int romanToInt(string s) {
unordered_map<char, int> roman;
roman['I'] = 1;
roman['V'] = 5;
roman['X'] = 10;
roman['L'] = 50;
roman['C'] = 100;
roman['D'] = 500;
roman['M'] = 1000;
int i = 0;
int r = 0;
int c = 0;
while (i<s.size()) {
if (i < s.size() - 1 && roman[s[i]] < roman[s[i+1]]) {
r += (roman[s[i+1]] - roman[s[i]]);
i+=2;
continue;
}
c = roman[s[i]];
i ++;
while (i < s.size() &&roman[s[i]] == roman[s[i-1]]) {
c += roman[s[i]];
i++;
}
r += c;
}
return r;
}
};