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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

思路：
罗马数字转成阿拉伯数字，这里面需要知道罗马数字的构成规则。罗马数字通过7个不同字母的重复或组合，能够表示出所有正整数（罗马数字中没有0）。
I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1000
比如：IV表示4，VI表示6，XIX表示19，XXI表示21。

可以找到规律，如果左边的字母表示的数字小于右边的字母，则用右边的数字减去左边的数字；反之，则需要进行加法。

我们从左向右进行遍历的时候，不太好计算出最终值；从另一个角度考虑，就是从右向左计算，就可以看出来了。

public class Solution {
    private static Map<Character, Integer> map = new HashMap() {
        {
            put('I', 1);
            put('V', 5);
            put('X', 10);
            put('L', 50);
            put('C', 100);
            put('D', 500);
            put('M', 1000);
        }
    };

    public static int romanToInt(String s) {
        int length = s.length();
        int result = 0;
        int preVal = 0;
        for (int i = length - 1; i >= 0; i--) {
            char key = s.charAt(i);
            int value = map.get(key);
            if (value >= preVal) {
                result += value;
            } else {
                result -= value;
            }

            preVal = value;
        }
        return result;
    }
}