lintcode-罗马数字转整数-419

class Solution {
public:
    
    int romanToInt(string& s) {
      
        if(s.size()==0)
            return 0;
        map<char,int>  cnt;
        map<char,bool> check;
        check['I']=true;
        check['X']=true;
        check['C']=true;
        int  sum=0;
        int  size=s.size();
        cnt['I']=1;;
        cnt['X']=10;
        cnt['C']=100;
        cnt['M']=1000;
        cnt['V']=5;
        cnt['L']=50;
        cnt['D']=500;
        
        for(int i=0;i<size;){
            if(i+1<size&&check[s[i]]&&cnt[s[i]]<cnt[s[i+1]]){
                sum+=cnt[s[i+1]]-cnt[s[i]];
                i+=2;
            }else{
                sum+=cnt[s[i]];
                ++i;
            }    
        }
        return sum;    
    }
};