(leisure) leetcode13 Roman to Integer

Topic link ：: delivery
Learn something ：

1. C I directly traverse the language , Actually, it's written later , I found that I can actually write a function , It will be simpler to write in this way .
2. Python The length of the first traversal side is 2 Of , Then go over it again, and the length is 1 Of ; You can use Python Dictionary in , then ,in and not in useful .

Code ：
Python:

class Solution:
    def romanToInt(self, s: str) -> int:
        my_bo = []; num = 0;
        my_dictionary = {
    
            'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000,
            'IV': 4, 'IX': 9, 'XL': 40, 'XC': 90, 'CD': 400, 'CM': 900,
        }

        for i in range(0, len(s) - 1):
            ss = s[i] + s[i + 1];
            if(ss in my_dictionary):
                num += my_dictionary[ss]
                my_bo.append(i); my_bo.append(i + 1);
                i += 1;
        
        for i in range(0, len(s)):
            if(i not in my_bo):
                num += my_dictionary[s[i]]
        
        return num
                

C++:

class Solution {
    
public:
    int romanToInt(string s) {
    
        string str = s;
        int sum = 0;
        for(int i = 0; i < s.size(); i ++) {
    
            if(str[i] == 'I') {
    
                if(i + 1 < s.size()) {
     
                    if(str[i + 1] == 'V') {
       //IV
                        sum += 4;
                        i ++;
                    }
                    else if(str[i + 1] == 'X') {
      //IX
                        sum += 9;
                        i ++;
                    } else sum += 1;  //I
                } else {
    
                    sum += 1;  //I
                }
                
            } else if(str[i] == 'V') {
    
                sum += 5;
            } else if(str[i] == 'X') {
    
                if(i + 1 < s.size()) {
     
                    if(str[i + 1] == 'L') {
       //XL
                        sum += 40;
                        i ++;
                    }
                    else if(str[i + 1] == 'C') {
      //LC
                        sum += 90;
                        i ++;
                    } else sum += 10;  //L
                } else {
    
                    sum += 10;  //L
                }
            } else if(str[i] == 'L') {
    
                sum += 50;
            } else if(str[i] == 'C') {
    
                if(i + 1 < s.size()) {
     
                    if(str[i + 1] == 'D') {
       //CD
                        sum += 400;
                        i ++;
                    }
                    else if(str[i + 1] == 'M') {
      //CM
                        sum += 900;
                        i ++;
                    } else sum += 100;  //C
                } else {
    
                    sum += 100;  //C
                }
            } else if(str[i] == 'D') {
    
                sum += 500;
            } else if(str[i] == 'M') {
    
                sum += 1000;
            }
        }

        return sum;
    }
};