13. Roman to Integer

LeetCode
,

My Solution (Greedy)

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class Solution {
    public int romanToInt(String s) {
        int result = 0;
        int pos = 0;
        int max = s.length();
        while (pos < max){
            if (s.charAt(pos) == 'M'){
                result += 1000;
                pos++;
            } else if (s.charAt(pos) == 'D'){
               result += 500;
               pos++;
            } else if (s.charAt(pos) == 'C'){
                if ((pos < max - 1) && (s.charAt(pos + 1) == 'M')){
                    result += 900;
                    pos += 2;
                } 
                else if ((pos < max - 1) && (s.charAt(pos + 1) == 'D')){
                    result += 400;
                    pos += 2;
                } else{
                    result += 100;
                    pos++;
                }
            } else if (s.charAt(pos) == 'L'){
                result += 50;
                pos++;
            }  else if (s.charAt(pos) == 'I'){
                 if ((pos < max - 1) && (s.charAt(pos + 1) == 'V')){
                    result += 4;
                    pos += 2;
                } else if ((pos < max - 1) && (s.charAt(pos + 1) == 'X')){
                    result += 9;
                    pos += 2;
                } else {
                    result += 1;
                    pos++;
                }
            } else if (s.charAt(pos) == 'X'){
                if ((pos < max - 1) && (s.charAt(pos + 1) == 'C')){
                    result += 90;
                    pos += 2;
                } 
                else if ((pos < max - 1) && (s.charAt(pos + 1) == 'L')){
                    result += 40;
                    pos += 2;
                } else{
                    result += 10;
                    pos++;
                }
            } else if (s.charAt(pos) == 'V'){
                    result += 5;
                    pos++;
            }
        }
        return result;
    }
}

Formatted

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class Solution {
    public int romanToInt(String s) {
        int result = 0, i = 0, n = s.length();
        while (i < n) {
            char c = s.charAt(i);
            if (c == 'I' && i + 1 < n && s.charAt(i + 1) == 'V') { result += 4; i += 2; }
            else if (c == 'I' && i + 1 < n && s.charAt(i + 1) == 'X') { result += 9; i += 2; }
            else if (c == 'X' && i + 1 < n && s.charAt(i + 1) == 'L') { result += 40; i += 2; }
            else if (c == 'X' && i + 1 < n && s.charAt(i + 1) == 'C') { result += 90; i += 2; }
            else if (c == 'C' && i + 1 < n && s.charAt(i + 1) == 'D') { result += 400; i += 2; }
            else if (c == 'C' && i + 1 < n && s.charAt(i + 1) == 'M') { result += 900; i += 2; }
            else {
                if (c == 'I') result += 1;
                else if (c == 'V') result += 5;
                else if (c == 'X') result += 10;
                else if (c == 'L') result += 50;
                else if (c == 'C') result += 100;
                else if (c == 'D') result += 500;
                else if (c == 'M') result += 1000;
                i++;
            }
        }
        return result;
    }
}

Remarks:

  1. Hard coded is faster than HashMap, due to reduced overhead from object instantiation, string slicing, and hash lookups