# 13. Roman to Integer

## Description

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.
``````

Tags: Math, String

## 题解

### 思路1

• 相同的数字连写，所表示的数等于这些数字相加得到的数，如 Ⅲ=3；

• 小的数字在大的数字的右边，所表示的数等于这些数字相加得到的数，如 Ⅷ=8、Ⅻ=12；

• 小的数字（限于 Ⅰ、X 和 C）在大的数字的左边，所表示的数等于大数减小数得到的数，如 Ⅳ=4、Ⅸ=9。

``````class Solution {
public int romanToInt(String s) {
Map<Character, Integer> map = new HashMap<>();
map.put('I', 1);
map.put('V', 5);
map.put('X', 10);
map.put('L', 50);
map.put('C', 100);
map.put('D', 500);
map.put('M', 1000);
int len = s.length();
int sum = map.get(s.charAt(len - 1));
for (int i = len - 2; i >= 0; --i) {
if (map.get(s.charAt(i)) < map.get(s.charAt(i + 1))) {
sum -= map.get(s.charAt(i));
} else {
sum += map.get(s.charAt(i));
}
}
return sum;
}
}
``````