LeetCode Roman to Integer

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Given a roman numeral, convert it to an integer.

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

这个题按照罗马数字的计数方法直接模拟就好，关于罗马数字的计数方法可以参考百度百科：http://baike.baidu.com/link?url=uRFEbVjfAx0doZf3O_f7-ur2vss1PnmMwdbONVVF6DKZpEegXlcL0IdQFqZnvnv2MtqLx32YZ3p1-JmhtM7MP_

主要看这三条规则就好：

V 和 X 左边的小数字只能用 Ⅰ；
L 和 C 左边的小数字只能用X；
D 和 M 左边的小数字只能用 C。

class Solution {public:    int romanToInt(string s) {        int h[257] = {0};        h['I'] = 1;        h['V'] = 5;        h['X'] = 10;        h['L'] = 50;        h['C'] = 100;        h['D'] = 500;        h['M'] = 1000;        int sum = 0;        for(int i = 0; i < s.size(); i++){            if(s[i] == 'I'){                if(i+1 < s.size() && (s[i+1] == 'X' || s[i+1] == 'V')){                    sum = sum + h[s[i+1]] - h[s[i]];                    i++;                }                else                    sum += h[s[i]];            }            else if(s[i] == 'X'){                if(i+1 < s.size() && (s[i+1] == 'L' || s[i+1] == 'C')){                    sum = sum + h[s[i+1]] - h[s[i]];                    i++;                }                else                    sum += h[s[i]];            }            else if(s[i] == 'C'){                if(i+1 < s.size() && (s[i+1] == 'D' || s[i+1] == 'M')){                    sum = sum + h[s[i+1]] - h[s[i]];                    i++;                }                else                    sum += h[s[i]];            }            else                sum += h[s[i]];        }                return sum;    }};

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