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RomanToInteger.java
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RomanToInteger.java
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package by.andd3dfx.numeric;
import java.util.Map;
/**
* <pre>
* <a href="https://leetcode.com/problems/roman-to-integer/description/">Task description</a>
*
* 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].
* </pre>
*
* @see <a href="https://youtu.be/vlKjGPFyltU">Video solution</a>
*/
public class RomanToInteger {
private static final Map<Character, Integer> vocabulary = Map.of(
'I', 1,
'V', 5,
'X', 10,
'L', 50,
'C', 100,
'D', 500,
'M', 1000
);
public static int romanToInt(String s) {
var sum = 0;
var pos = 0;
while (pos < s.length()) {
var left = vocabulary.get(s.charAt(pos));
if (pos == s.length() - 1) {
sum += left;
break;
}
var right = vocabulary.get(s.charAt(pos + 1));
if (left >= right) {
sum += left;
} else {
sum -= left;
}
pos++;
}
return sum;
}
}