1 class Solution(object): 2     def intToRoman(self, num): 3         """ 4         :type num: int 5         :rtype: str 6         """ 7         rom = {
   'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000} 8          9         ans = []10         p = num/100011         ans += ['M']*p12         num = num%100013         14         p = num/10015         if p <= 3:16             ans += ['C']*p17         elif p == 4:18             ans += ['CD']19         elif 5<= p <= 8 :20             ans += ['D']+['C']*(p-5)21         elif p == 9:22             ans += ['CM']23         num = num%10024         25         p = num/1026         if p <= 3:27             ans += ['X']*p28         elif p == 4:29             ans += ['XL']30         elif 5 <= p <=8 :31             ans += ['L']+['X']*(p-5)32         elif p == 9:33             ans += ['XC']34         num = num%1035         36         p = num37         if p <= 3:38             ans += ['I']*p39         elif p == 4:40             ans += ['IV']41         elif 5 <= p <= 8:42             ans += ['V']+['I']*(p-5)43         elif p == 9:44             ans += ['IX']45         return ''.join(ans)46

 

Given a roman numeral, convert it to an integer.

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

 

思路：先建立一个对应的dict。注意转换规则中的特例，比如XL (=50-10), IV (=5-1), 也就是说当前的字母如果小于后面的那个，当前这个字母对应的数字取负。

1 class Solution(object): 2     def romanToInt(self, s): 3         """ 4         :type s: str 5         :rtype: int 6         """ 7         rNum = {
   'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000} 8         size = len(s) 9         if size == 1:10             return rNum[s[0]]11         sum = 012         for i in range(size):13             if i < size - 1 and rNum[s[i]] < rNum[s[i+1]]:14                 sum -= rNum[s[i]]15             else:16                 sum += rNum[s[i]]17         return sum