[leetcode]Roman to Integer @ Python

原题地址:https://oj.leetcode.com/problems/roman-to-integer/

题意:

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

解题思路:将罗马数字转换成对应的整数。首先将罗马数字翻转,从小的开始累加,如果遇到CM(M-C=1000-100=900)这种该怎么办呢?因为翻转过来是MC,M=1000先被累加,所以使用一个last变量,把M记录下来,如果下一个数小于M,那么减两次C,然后将C累加上,这个实现比较巧妙简洁。

代码:

class Solution:
    # @return an integer
    def romanToInt(self, s):
        numerals = { "M": 1000, "D": 500, "C": 100, "L": 50, "X": 10, "V": 5, "I": 1 }
        sum=0
        s=s[::-1]
        last=None
        for x in s:
            if last and numerals[x]<last:
                sum-=2*numerals[x]
            sum+=numerals[x]
            last=numerals[x]
        return sum

 

posted @ 2014-06-10 11:54  南郭子綦  阅读(5346)  评论(0编辑  收藏  举报