13.Roman to Integer (HashTable)

Given a roman numeral, convert it to an integer.

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

class Solution {
public:
    int romanToInt(string s) {
        char c[] = {'I','V','X','L','C','D','M'};
        int n[] = {,,,,,,};
        unordered_map<char,int> map;
        map.insert(make_pair('I',));
        map.insert(make_pair('V',));
        map.insert(make_pair('X',));
        map.insert(make_pair('L',));
        map.insert(make_pair('C',));
        map.insert(make_pair('D',));
        map.insert(make_pair('M',));

        int len = s.length();
        int num = ;
        if(len==){
            num = map[s[]];
            return num;
        }

        for(int i = ; i < len; i++){
            if(map[s[i]] <= map[s[i-]]){
                num += map[s[i-]];
            }
            else{
                num+= map[s[i]]-map[s[i-]];
                i++; //dealt two letters at one time, so i should increase 2
            }
        }

        //Do not forget to discuss the last case independently
        if(len >  && map[s[len-]] <= map[s[len-]]){
            num+=map[s[len-]];
        }

        return num;
    }
};

Improve: 使用两个数组代替map，更节省空间。

class Solution {
public:
    int romanToInt(string s) {
        int values[] = {, , , , , ,  };
        char numerals[] = {'M', 'D', 'C', 'L', 'X', 'V', 'I' };
        int i = , j = , result = ;
        while(i < s.length()){
            if(s[i] != numerals[j]){
                j++;
                continue;
            }

            if(i+<s.length() && j->= && s[i+] == numerals[j-]){
                result += values[j-]-values[j];
                i+=;
            }
            else if(i+<s.length() && j->= && s[i+] == numerals[j-]){
                result += values[j-]-values[j];
                i+=;
            }
            else{
                result += values[j];
                i++;
            }
        }
        return result;
    }
};