Roman to Integer [LeetCode]

Given a roman numeral, convert it to an integer.

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

Summary: When meeting C/X/I, remembers to search forward to check if there is a bigger number at the front.

     int romanToInt(string s) {
         if(s.size() == )
             return ;

         int num = ;
         int m_idx = -;
         int d_idx = -;
         for(int i = ; i < s.size(); i ++) {
             if(s[i] == 'C') {
                 m_idx = s.find_first_of('M', i + );
                 d_idx = s.find_first_of('D', i + );
                 if(m_idx > i && m_idx < s.size()){
                    num +=  - (m_idx - i)*;
                    i = m_idx;
                 }else if(d_idx > i && d_idx < s.size()){
                    num +=  - (d_idx - i)*;
                    i = d_idx;
                 }else {
                     num += ;
                 }
             }else if( s[i] == 'M'){
                 num += ;
             }else if(s[i] == 'D') {
                 num += ;
             }else if(s[i] == 'L') {
                 num += ;
             }else if(s[i] == 'X') {
                 m_idx = s.find_first_of('C', i + );
                 d_idx = s.find_first_of('L', i + );
                 if(m_idx > i && m_idx < s.size()){
                    num +=  - (m_idx - i)*;
                    i = m_idx;
                 }else if(d_idx > i && d_idx < s.size()){
                    num +=  - (d_idx - i)*;
                    i = d_idx;
                 }else {
                     num += ;
                 }
             }else if(s[i] == 'V'){
                 num += ;
             }else if(s[i] == 'I') {
                 m_idx = s.find_first_of('X', i + );
                 d_idx = s.find_first_of('V', i + );
                 if(m_idx > i && m_idx < s.size()){
                    num +=  - (m_idx - i);
                    i = m_idx;
                 }else if(d_idx > i && d_idx < s.size()){
                    num +=  - (d_idx - i);
                    i = d_idx;
                 }else {
                     num += ;
                 }
             }
         }
         return num;
     }