Leetcode-Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

Solution:

public class Solution {
    public int romanToInt(String s) {
        if (s.length()==0) return 0;

        Map<Character,Integer> map = new HashMap<Character,Integer>();
        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);
        
        int res = 0;
        int index = 0;
        while (index<s.length()){
            char cur = s.charAt(index);
            if (index+1<s.length()){
                char next = s.charAt(index+1);
                int val1 = map.get(cur);
                int val2 = map.get(next);
                if (val1<val2){
                    res += (val2-val1);
                    index += 2;
                } else {
                    res += val1;
                    index++;
                }
            } else {
                int val = map.get(cur);
                res += val;
                index++;
            }
        }

        return res;       
    }
}

 Solution 2:

Whenever the value of char i is smaller than the value of char i+1, then the value of char i should be reduced from the sum.

public class Solution {
    public int romanToInt(String s) {
        Map<Character,Integer> map = new HashMap<Character,Integer>();
        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);
       
        int sum = 0;
        for (int i=0;i<s.length()-1;i++){
            char c = s.charAt(i);
            int val = map.get(c);
            if (c=='I' || c=='X' || c=='C'){
                char next = s.charAt(i+1);
                int val2 = map.get(next);
                if (val<val2) val = -val;
            }

            sum += val;
        }
        char c = s.charAt(s.length()-1);
        sum += map.get(c);
        return sum;
    }
}