LeetCode——13. Roman to Integer

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博客介绍了罗马数字的表示规则，包括基本符号及其对应值，以及减法规则。给出了将罗马数字转换为整数的问题，提供多个示例展示转换结果。还给出解题思路，可从后往前遍历，结合字典存储对应关系进行转换，输入范围在1到3999。

LeetCode——13. Roman to Integer

Description
Examples
Note
Solution

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.

Examples

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.

Note

可以从后往前看，如何后面的罗马数字比前面的大，则将当前的数字前去前一个数字，如果当前的数字比后面的数字大，则当前的数字加上当前罗马数字所代表的数字。
可以将罗马数字与整数值之间的对应关系设置为一个字典，将字符串s从后往前便利。

Solution

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        
        list = {
            
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000 
        }
        
        value = 0
        current_value = 0
        for item in s[::-1]:
            temp = list[item]
            if list[item] >= current_value:
                value += list[item]
                current_value = list[item]
            else:
                value = value - list[item]
        return value