## Monday, December 31, 2012

### [LeetCode] Roman To Integer 解题报告

Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
» Solve this problem

[解题思路]

• 如果当前比前一个大，说明这一段的值应该是当前这个值减去上一个值。比如IV = 5 – 1
• 否则，将当前值加入到结果中，然后开始下一段记录。比如VI = 5 + 1， II=1+1

[Code]
``````1:    inline int c2n(char c) {
2:      switch(c) {
3:        case 'I': return 1;
4:        case 'V': return 5;
5:        case 'X': return 10;
6:        case 'L': return 50;
7:        case 'C': return 100;
8:        case 'D': return 500;
9:        case 'M': return 1000;
10:        default: return 0;
11:      }
12:    }
13:    int romanToInt(string s) {
14:      // Start typing your C/C++ solution below
15:      // DO NOT write int main() function
16:      int result=0;
17:      for(int i =0; i< s.size(); i++)
18:      {
19:        if(i>0&& c2n(s[i]) > c2n(s[i-1]))
20:        {
21:          result +=(c2n(s[i]) - 2*c2n(s[i-1]));
22:        }
23:        else
24:        {
25:          result += c2n(s[i]);
26:        }
27:      }
28:      return result;
29:    }
``````

http://fisherlei.blogspot.com/2012/12/leetcode-integer-to-roman.html