【leetcode】Roman to Integer

Question ：

Given a roman numeral, convert it to an integer.

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

Anwser 1 ：

class Solution {
public:
    int toNum(char c)  
    {  
        switch(c)  
        {  
            case 'I':  
                return 1;  
            case 'V':  
                return 5;  
            case 'X':  
                return 10;  
            case 'L':  
                return 50;  
            case 'C':  
                return 100;  
            case 'D':  
                return 500;  
            case 'M':  
                return 1000;  
            default:  
                return 0;  
        }  
    } 
    
    int romanToInt(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ret = 0;
    	for(int i = 0; i < s.size(); i++)
		{
			if(i + 1 < s.size())
			{
				if(toNum(s[i]) < toNum(s[i+1]))     // high num < low num, then minus
				{
					ret -= toNum(s[i]);
				}
				else
				{
					ret += toNum(s[i]);
				}
			}
			else
			{
				ret += toNum(s[i]);
			}
		}
		return ret;
    }
};

Anwser 2 ：

class Solution {
public:
    int toNum(char c)  
    {  
        switch(c)  
        {  
            case 'I':  
                return 1;  
            case 'V':  
                return 5;  
            case 'X':  
                return 10;  
            case 'L':  
                return 50;  
            case 'C':  
                return 100;  
            case 'D':  
                return 500;  
            case 'M':  
                return 1000;  
            default:  
                return 0;  
        }  
    } 
    
    int romanToInt(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ret = 0;
        
        int len = s.size();
        int pre = 1001;
        int cur;
        for (int i = 0; i < len; ++i)
        {
            cur = toNum(s[i]);
            ret += cur;
            if (cur > pre)
            {
                ret -= 2*pre;   // pre already add one time, now minus tow times
            }
            pre = cur;
        }
        return ret;
    }
};