[Algorithm] Leetcode - 13. Roman to Integer (JAVA)

Roman to Integer

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 one’s 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.

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].

Example

#1

Input: s = "III"
Output: 3
Explanation: III = 3.

#2

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

#3

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

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내가 작성한 코드

처음엔 indexOf 로 위치 비교도 했다가 switch 도 썼다가 결국 아래와 같이 HashMap 을 이용하여 작성했다.

먼저 HashMap에 문자열과 숫자 값을 세팅해준다.
그리고 for 반복문을 이용하여 문자열의 맨 오른쪽에서 두 번째부터 반복하여 비교해준다. 앞의 문자열이 뒤의 문자열보다 숫자 값이 크거나 같을 경우 더해주고, 작을 경우에는 빼준다.

class Solution {
    public int romanToInt(String s) {
        int result;
        HashMap<Character, Integer> map = new HashMap<>();
        
        map.put('I', 1);
        map.put('V', 5);
        map.put('X', 10);
        map.put('L', 50);
        map.put('C', 100);
        map.put('D', 500);
        map.put('M', 1000);
        
        result = map.get(s.charAt(s.length() - 1));
        
        for (int i = s.length() - 2; i >= 0; i--) {
            if (map.get(s.charAt(i)) >= map.get(s.charAt(i + 1))) {
                result += map.get(s.charAt(i));
            } else {
                result -= map.get(s.charAt(i));
            }
        }
        
        return result;
    }
}

속도 면에서 더 빠른 코드가 있어서 가져와 보았다. 아래 코드는 switch 문을 이용하여 풀이했다.

class Solution {
    public int romanToInt(String s) {
        int sum = 0;
        char prev = '?';
        
        for(int i = s.length()-1; i >= 0; i--)    {
            char curr = s.charAt(i);
            switch(curr)    {
                case 'I'    : sum += prev == 'V' || prev == 'X' ? -1 : 1;
                    break;
                case 'V'    : sum += 5;
                    break;
                case 'X'    : sum += prev == 'L' || prev == 'C' ? -10 : 10;
                    break;
                case 'L'    : sum += 50;
                    break;
                case 'C'    : sum += prev == 'D' || prev == 'M' ? -100 : 100;
                    break;
                case 'D'    : sum += 500;
                    break;
                case 'M'    : sum += 1000;
                    break;
                default     : 
                    break;
                }
                prev = curr;
            }
        return sum;        
    }
}

더 많은 알고리즘 문제 풀이는 여기서 확인할 수 있다.