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. 

 1     int romanToInt(string s) {

 2         if(s.size() == 0)

 3             return 0;

 4             

 5         int num = 0;

 6         int m_idx = -1;

 7         int d_idx = -1;

 8         for(int i = 0; i < s.size(); i ++) {

 9             if(s[i] == 'C') {

10                 m_idx = s.find_first_of('M', i + 1);

11                 d_idx = s.find_first_of('D', i + 1);

12                 if(m_idx > i && m_idx < s.size()){

13                    num += 1000 - (m_idx - i)*100; 

14                    i = m_idx;

15                 }else if(d_idx > i && d_idx < s.size()){

16                    num += 500 - (d_idx - i)*100; 

17                    i = d_idx;

18                 }else {

19                     num += 100;

20                 }

21             }else if( s[i] == 'M'){

22                 num += 1000;

23             }else if(s[i] == 'D') {

24                 num += 500;

25             }else if(s[i] == 'L') {

26                 num += 50;

27             }else if(s[i] == 'X') {

28                 m_idx = s.find_first_of('C', i + 1);

29                 d_idx = s.find_first_of('L', i + 1);

30                 if(m_idx > i && m_idx < s.size()){

31                    num += 100 - (m_idx - i)*10; 

32                    i = m_idx;

33                 }else if(d_idx > i && d_idx < s.size()){

34                    num += 50 - (d_idx - i)*10; 

35                    i = d_idx;

36                 }else {

37                     num += 10;

38                 }

39             }else if(s[i] == 'V'){

40                 num += 5;

41             }else if(s[i] == 'I') {

42                 m_idx = s.find_first_of('X', i + 1);

43                 d_idx = s.find_first_of('V', i + 1);

44                 if(m_idx > i && m_idx < s.size()){

45                    num += 10 - (m_idx - i);

46                    i = m_idx;

47                 }else if(d_idx > i && d_idx < s.size()){

48                    num += 5 - (d_idx - i); 

49                    i = d_idx;

50                 }else {

51                     num += 1;

52                 }

53             }

54         }

55         return num;

56     }