Leetcode - 13. Roman to Integer

문제

로마자 표기를 10진수 값으로 리턴하기.
https://leetcode.com/problems/roman-to-integer/

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

• 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.

해결

1. Constraints
 - IV: 4
 - IX: 9
 - XL: 40
 - XC: 90
 - CD: 400
 - CM: 900

2. Ideas
 - calculate linear ch by ch, but some char(I,X,C) check next char. -> O(N)

3. Test Cases
 - assert(romanToInt("LVIII") == 58)
 - assert(romanToInt("IV") == 4)
 - assert(romanToInt("VIII") == 8)
 - assert(romanToInt("IX") == 9)

4. Coding

int romanToInt(char * s){
    int ssize = strlen(s);
    int ret = 0;
    for (int i = 0; i < ssize; i++) {
        switch (s[i]) {
            case 'I':
                if (i + 1 < ssize && s[i + 1] == 'V') {
                    ret += 4;    
                    i++;
                    continue;
                }
                if (i + 1 < ssize && s[i + 1] == 'X') {
                    ret += 9;    
                    i++;
                    continue;
                }
                ret += 1;
                break;
            case 'V':
                ret += 5;
                break;
            case 'X':
                if (i + 1 < ssize && s[i + 1] == 'L') {
                    ret += 40;    
                    i++;
                    continue;
                }
                if (i + 1 < ssize && s[i + 1] == 'C') {
                    ret += 90;    
                    i++;
                    continue;
                }
                ret += 10;
                break;
            case 'L':
                ret += 50;
                break;
            case 'C':
                if (i + 1 < ssize && s[i + 1] == 'D') {
                    ret += 400;    
                    i++;
                    continue;
                }
                if (i + 1 < ssize && s[i + 1] == 'M') {
                    ret += 900;    
                    i++;
                    continue;
                }
                ret += 100;
                break;
            case 'D':
                ret += 500;
                break;
            case 'M':
                ret += 1000;
                break;
        }
    }
    return ret;
}