【leetcode】13. Roman to Integer

罗马数字转换整数解析

最新推荐文章于 2023-10-30 12:20:17 发布

jinjinjin2014

最新推荐文章于 2023-10-30 12:20:17 发布

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本文详细解析了LeetCode上罗马数字转整数的问题，介绍了罗马数字的基本规则及特殊情况处理方法，并提供了一种有效的解决方案。

一、题目描述(罗马数字转为数字)

https://leetcode.com/problems/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.

二、题目分析

发现规律：罗马数字为从大到小排列，若前面的罗马数字小于后面，则此两位相当于减去前面数字再加上后面数字，按照这个规律可解。
注意出错处：
1.由于range函数生成的range是一个范围，它是左闭右开区间，所以循环次数需要注意。
2.循环中有s[i+1]出现，应该保证s[i+1]有意义，所以在for循环中少循环一次，s最后一项的值在循环外加上，即sum+=d[s[-1]]

三、代码实现

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