LeetCode: Roman to Integer

最新推荐文章于 2025-08-06 17:01:48 发布

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最新推荐文章于 2025-08-06 17:01:48 发布

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分类专栏： LeetCode 文章标签： LeetCode 算法 Python

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本文详细介绍了罗马数字的基本符号及其对应的数值，解释了罗马数字的书写规则，并提供了一个将罗马数字转换为整数的Python代码示例。通过实例演示，如III、IV、IX、LVIII和MCMXCIV等罗马数字的转换过程，帮助读者理解和掌握罗马数字的计算方法。

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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 an integer, convert it to a roman numeral. 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.

与上一题相反，本题要求将罗马数字转换为int。为了方便起见，我将罗马数字代表的int用一个dictionary列出来，并将res初始化为d[s[0]]，即第一个字符代表的数字。

如果s长度为1，则直接返回结果
如果s长度大于1，则遍历s[1:]，将每个字符代表的int加到res里，如果某一个字符代表的int大于前一个字符代表的int，则应该将res减去前一个字符代表的int值的两倍
返回res

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        d = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        res = d[s[0]]
        if len(s) == 1:
            return res
        for i, v in enumerate(s[1:]):
            res += d[v]
            if d[v] > d[s[i]]:
                res -= d[s[i]] * 2
        return res