题目：

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

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

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

描述：

给出一个罗马数字，转化为阿拉伯数字后输出

 

分析：

将每种罗马字符对应的阿拉伯数字存储到一张表中，从第一个字符依次累加遍历到的字符所对应的阿拉伯数值，其中在遇到后一个字符表示的数值大于前一个字符表示的数值时，则这两个字符表示的是较大字符表示的数值减去较小字符所表示的数值：比如IV和IX。

 

代码：（时间复杂度O（n））

class Solution {
public:
	int sign[128];
	void init() {
		sign['I'] = 1;
		sign['V'] = 5;
		sign['X'] = 10;
		sign['L'] = 50;
		sign['C'] = 100;
		sign['D'] = 500;
		sign['M'] = 1000;
	}
	
    int romanToInt(string s)  {
    	init();
        int result = 0;
        for (int i = 0; i < s.size(); ++ i) {
        	int value = sign[s.at(i)];
        	if (i + 1 < s.size() && sign[s.at(i)] < sign[s.at(i + 1)]) {
        		value =  sign[s.at(i + 1)] - sign[s.at(i)];
        		++ i;
			}
			result += value;
		} 
		return result;
    }
};