Leetcode 013 罗马转数字思路详解+反思总结 Pyt

最新推荐文章于 2021-12-12 17:33:56 发布

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最新推荐文章于 2021-12-12 17:33:56 发布

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分类专栏： Python学习 Leetcode 刷题记录python,C++, JAVA

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本文介绍了一种将罗马数字转换为整数的方法，通过预先处理特殊情况简化问题，并提供了一个具体的Python实现示例。

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.

思路：因为这道题4和9的特殊性，于是我就想到了先把特殊的字符先替换掉的操作，于是剩下的罗马数字都是单个字符的了。

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        data = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        result = 0
        if 'CD' in s:
            result += 400
            s = s.replace("CD", "")
        elif 'CM' in s:
            result += 900
            s = s.replace("CM", "")
        if 'XL' in s:
            result += 40
            s = s.replace("XL", "")
        elif 'XC' in s:
            result += 90
            s = s.replace("XC", "")
        if 'IV' in s:
            result += 4
            s = s.replace("IV", "")
        elif 'IX' in s:
            result += 9
            s = s.replace("IX", "")
        for i in set(s):
            result += data[i]*(s.count(i))
        return result

Beat 98

总结：遇到难处理的特殊情况，不如把特殊情况给消除了，剩下的就是正常情况。