[LeetCode] 13. Roman to Integer (Java)

Description

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.

 

주어진 로마자를 정수로 변환하기.

 

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

class Solution {
    public int romanToInt(String s) {
        int now = 0, bef = 1001, ans = 0;
        
        for(int i=0;i<s.length();i++){
            switch(s.charAt(i)){
                case 'I':
                    now = 1;
                    break;
                case 'V':
                    now = 5;
                    break;
                case 'X':
                    now = 10;
                    break;
                case 'L':
                    now = 50;
                    break;
                case 'C':
                    now = 100;
                    break;
                case 'D':
                    now = 500;
                    break;
                case 'M':
                    now = 1000;
                    break;
            }
            if(now>bef){
                ans += now-(2*bef);
                bef = now;
            }else{
                ans += now;
                bef = now;
            }
        }
        
        return ans;
    }
}

오랜만에 switch문을 사용해서 조금 버벅였지만 문제의 풀이 자체는 이전 문제와 비슷하게 간단했다.