1 Description（描述）
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.
罗马数字通过其中不同的符号来表示：I, V, X, L, C, D 和 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.
举个例子，2用罗马数字表示为 II，仅仅将两个 I 加到一起。12写成 XII 。27写成 XXVII。
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.
罗马数字通常都是从大到小、从左到右。然而，罗马数字表示4不是 IIII 。反而，4写为 IV 。因为1在5之前，我们减掉他变成4。相同的道理适合9，它表示成 IX。使用减法的例子有六种：
- 把 I 放到 V 和 X 之前可以表示 4 和 9.
- 把 X 放到 L 和 C 之前可以表示 40 和 90.
- 把 C 放到 D 和 M 之前可以表示 400 和 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
给定一个罗马数字，将它转化成一个整数。输入值在1~3999。
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.
2 Solution（解决方案）
官网没有给标准答案，以下是我的代码。
public static int romanToInt(String s){HashMap<Character,Integer> hash = new HashMap<Character,Integer>(){
{put('I', 1);put('V', 5);put('X', 10);put('L', 50);put('C', 100);put('D', 500);put('M', 1000);}};char romans[] = s.toCharArray();int sum = 0;for(int i = 0; i < romans.length - 1; i++){int temp1 = hash.get(romans[i]);int temp2 = hash.get(romans[i+1]);if(temp1 < temp2 ){sum -= temp1;}else{sum += temp1;}}return sum + hash.get(romans[romans.length - 1]);
}