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.
Input: s = "III"
Output: 3
Explanation: III = 3.
Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 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].
Contents
- Solution 1 - Using the order of Roman symbols
In this approach, we will iterate through the characters of input string s, for the current character at index i
check if the integer value of next character at index i+1 is greater than
current character or not, if it is greater then we can subtract that from the result, otherwise we can add it to the result. (we are guaranteed that the input string will be
a valid roman integer, so we dont need to check if the smaller value came before the specific characters mentioned in problem statement)
import java.util.Map;
public class RomanToInteger {
static final Map<Character, Integer> roman = Map.of('I',1,
'V', 5,
'X',10,
'L',50,
'C',100,
'D',500,
'M',1000);
static int romanToInt(String s) {
int result =0;
for (int i=0; i<s.length(); i++) {
int current = roman.get(s.charAt(i));
if(i+1 < s.length() && current < roman.get(s.charAt(i+1))) {
// if there is next character, check if current is lesser than next one
result -= current;
} else {
result += current;
}
}
return result;
}
public static void main(String[] args) {
System.out.println(romanToInt("III"));
System.out.println(romanToInt("LVIII"));
System.out.println(romanToInt("MCMXCIV"));
}
}
Complexity Analysis:
Time complexity: Above code runs in O(n) time where n is the length of input string s.
Space complexity: O(1).
Above implementations source code can be found at
GitHub link for Java code