---
title: 【LeetCode】0013. Roman to Integer
date: 2019-06-06
is_modified: false
disqus: cynthiahackmd
categories:
- "面試刷題"
tags:
- "LeetCode"
---

{%hackmd @CynthiaChuang/Github-Page-Theme %}

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

<!--more-->
|**Symbol**|**Value**|
|-------------|---|
|I|1|
|V|5|
|X|10|
|L|50|
|C|100|
|D|500|
|M|1000|

<br>

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`.
<br>

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:
<br>

- `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.

<br>

**Related Topics:** `Math`、`String`

## 解題邏輯與實作
在羅馬數字的規則中有一條，若是左邊的數字比右邊小，則這個位數的值為右減左。其他情況下則需要把羅馬數至相對應的值相加。

但在實做時，若發現當前的數字比前一個數字大，才回頭更新值，會使的程式流程變得相當複雜，因此這邊將傳入的字串反過來操作。

class Solution:
    def romanToInt(self, s: str) -> int:
         roman_to_int = {"M": 1000, "D": 500, "C": 100, "L": 50, "X": 10, "V": 5, "I": 1}
         result = 0
         pre_int = 0
         for idx , char in enumerate(s[::-1]):
             now_int = roman_to_int [char]
             result += now_int * -1 if now_int < pre_int else now_int
             pre_int = now_int
			
         return result

## 其他連結
1. [【LeetCode】0000. 解題目錄](/x62skqpKStKMxRepJ6iqQQ)

<br><br>

> **本文作者**： 辛西亞．Cynthia
> **本文連結**： [辛西亞的技能樹](https://cynthiachuang.github.io/LeetCode-0013-Roman-to-Integer) / [hackmd 版本](https://hackmd.io/@CynthiaChuang/LeetCode-0013-Roman-to-Integer)
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