0013. Roman to Integer

# 13. Roman to Integer#

## 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.
``````

## 题目大意 #

``````

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

``````

• I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
• X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。
• C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。

## 解题思路 #

• 给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。
• 简单题。按照题目中罗马数字的字符数值，计算出对应罗马数字的十进制数即可。

## 代码 #

``````
package leetcode

var roman = map[string]int{
"I": 1,
"V": 5,
"X": 10,
"L": 50,
"C": 100,
"D": 500,
"M": 1000,
}

func romanToInt(s string) int {
if s == "" {
return 0
}
num, lastint, total := 0, 0, 0
for i := 0; i < len(s); i++ {
char := s[len(s)-(i+1) : len(s)-i]
num = roman[char]
if num < lastint {
total = total - num
} else {
total = total + num
}
lastint = num
}