Roman numerals contain the following seven characters: I, V, X, L, C, D, and M.

Character value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000
 For example, the Roman numeral 2 is written as "II", that is, two parallel ones. 12 write XII, that is, X + II. 27 write XXVII, that is, XX + V + II.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

*How to solve the problem: judge whether the number represented by the previous digit is smaller than the position, and subtract the number represented by the previous digit if it is smaller
 *(Note: since traversing the previous bit is adding the previous bit to res, you need to subtract twice the number represented by the previous bit)

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