Roman to Integer

題目：
Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.
思路：
我們首先來明確一下羅馬數字與阿拉伯數字的換算規則：如果當前數字比前一個大，說明這一段的值應該是當前這個值減去上一個值，比如IV = 5 – 1；否則，將當前值加入到結果中，然後開始下一段記錄，比如VI = 5 + 1, II=1+1。而羅馬數字與阿拉伯數字對應變換是：I對應1，V對應5，X對應10，L對應50，C對應100，D對應500，M對應1000。因此，只需要從前往後讀出字符，如果當前數字小於等於前一字符，則加上當前字符對應的數字；而當前數字更大時，減去前一個字符（要減去兩倍，因爲在前面掃描時已經加上過一次了，實際上不應該加，因此要多減一次）。
代碼：

class Solution {
public:
    int romanToInt(string s) {

char c[10][10][10]={{"0","I","II","III","IV","V","VI","VII","VIII","IX"},
                    {"0","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
                    {"0","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
                    {"0","M","MM","MMM"}};

         int num=0,i;

         for(i=0;i<s.size();i++)
         {
             if(s[i]=='I') num+=1;
             if(s[i]=='V') 
             {
                if(i!=0 && s[i-1]=='I') num+=3;
                else num+=5;
             }
             if(s[i]=='X')
             {
                 if(i!=0 && s[i-1]=='I') num+=8;
                 else num+=10;
             }
             if(s[i]=='L')
             {
                 if(i!=0 && s[i-1]=='X') num+=30;
                 else num+=50;
             }
             if(s[i]=='C')
             {
                 if(i!=0 && s[i-1]=='X') num+=80;
                 else num+=100;
             }
             if(s[i]=='D')
             {
                 if(i!=0 && s[i-1]=='C') num+=300;
                 else num+=500;
             }
             if(s[i]=='M')  
             {
                 if(i!=0 && s[i-1]=='C') num+=800;
                 else num+=1000;
             }
         }
         return num;
    }
};