5.Integer to Roman && Roman to Integer

Roman chart: http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm

Integer to Roman

Given an integer, convert it to a roman numeral.

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

解析：只有数字为 9 和 4 时，取一个小数放在左边。（把数字为 9 和 4 时的罗马符号看作一个原子，避免取小数过程。）

 class Solution {
 public:
     string intToRoman(int num) {
         string Roman[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
         int val[] = {, , , , , , , , , , , ,  };
         string s;
         for(int i = ; i < ; ++i){
             while(num >= val[i]){
                 num -= val[i];
                 s += Roman[i];
             }
         }
         return s;
     }
 };

 Roman to Integer

Given a roman numeral, convert it to an integer.

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

解析：1. 最直观想法，对字符串按罗马值从大到小比对。 (372 ms)

 bool inHead(string s1, string& s2)
 {
     if(s2 == "") return false;
     int k = ;
     auto it = s1.begin();
     for(; it != s1.end(); ++it)
     {
         if(*it != s2[k++]) return false;
     }
     if(it == s1.end()){
         s2.erase(, k);
         return true;
     }
 }

 class Solution {
 public:
     int romanToInt(string s) {
         string sigmal[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
         int a[] = {, , , , , , , , , , , ,  };
         int num = ;
         int k = ;
         for(int i = ; i < ; ++i){
             while(inHead(sigmal[i], s)){
                 num += a[i];
             }
         }
         return num;
     }
 };

Code

2. 利用 hash 函数来做。(每个罗马字符范围都在 'A' - 'Z' 之间，对字符串从左到右扫描，若是出现逆序（按罗马值），则这个逆序对为一个值) (260ms)

 class Solution {
 public:
     int romanToInt(string s) {
         char c[] = {'M', 'D', 'C', 'L', 'X', 'V', 'I'};
         int v[] = {, , , , , ,  };
         int hash[] = {};
         for(int i = ; i < ; ++i){
             hash[c[i]] = v[i];
         }
         int num = ;
         int len = s.length();
         for(int i = ; i < len; ++i){
             if(i == len-){
                 num += hash[s[i]];
                 return num;
             }
             if(hash[s[i]] < hash[s[i+]]){
                 num += hash[s[i+]] - hash[s[i]];
                 ++i;
             }else num += hash[s[i]];
         }
         return num;
     }
 };