# Convert roman numerals to Arabic numerals

Source: Internet
Author: User

` 1 /* I:1 ; V-5;X-10;L-50;C-100;D-500;M-1000*/ 2 class Solution{ 3     vector<string> tokens; 4     vector<char> token_value; 5     public: 6     /*divide the string into some tokens,every token including same char*/ 7     void token(string s) 8     { 9         string::iterator i=s.begin();10         string::iterator j=s.begin();11 12         while(i!=s.end())13         {14             char cur=*i;15             token_value.push_back(cur);16             while(*i == cur){17                 i++;18             }19 20             string temp(j,i);21             tokens.push_back(temp);22             j=i;23         }24     }25     /*比较相邻字母的大小*/26     bool lower(char a,char b)27     {28         if((a==‘I‘ && b!=‘I‘)||(a==‘V‘ && b!=‘I‘&&b!=‘V‘) || (a==‘X‘&&b!=‘I‘&&b!=‘V‘&&b!=‘X‘)|| 29                 (a==‘L‘&&b!=‘I‘&&b!=‘V‘&&b!=‘X‘&&b!=‘L‘)||(a==‘C‘ && (b==‘D‘ || b==‘M‘)) || (a==‘D‘&&b==‘M‘))30             return true;31         else return false;32     }33     int romanToInt(string s){34         token(s);35         int result=0;36         vector<int> res;37         /*将上述tokens转为一个int数组*/38         for(vector<string>::iterator i=tokens.begin();i!=tokens.end();i++)39         {40             string temp=*i;41             int temp_value=0;42             char cur=temp[0];43             int len=temp.size();44             for(int i=0;i<len;i++)45             {46                 switch(cur)47                 {48                     case ‘I‘:temp_value++;break;49                     case ‘V‘:temp_value+=5;break;50                     case ‘X‘:temp_value+=10;break;51                     case ‘L‘:temp_value+=50;break;52                     case ‘C‘:temp_value+=100;break;53                     case ‘D‘:temp_value+=500;break;54                     case ‘M‘:temp_value+=1000;break;55                     default:break;56                 }57 58             }59             res.push_back(temp_value);60 61         }62         result=res[res.size()-1];63         for(int i=token_value.size()-2;i>=0;i--)64         {65             if(lower(token_value[i],token_value[i+1])){66                 result-=res[i];67             }68             else{69                 result+=res[i];70             }71             cout<<result<<endl;72         }73         return result;74     }75 };1. 首先去维基百科查阅罗马数字的规范和表示方法；2. 由于做词法分析的缘故，对字符串的问题总是习惯分解成token来做，首先我将罗马数字按照其类别分成很多组，每一组内的元素都是相同的，然后将每组的值计算出来；3. 现在得到了2个新的数组，一个是由罗马数字类别组成的数组（可看成原字符串去掉重复之后的结果），另一个是上一个数组的结果数组；4. 最后我们根据第一个数组的char 的大小关系来对结果数组进行求值，从右向左求值，左边比右边大，就加上左边的，否则减去左边的值。`

Convert roman numerals to Arabic numerals

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