标签:style class blog code java http
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
[leetcode]Roman to Integer @ Python,布布扣,bubuko.com
[leetcode]Roman to Integer @ Python
标签:style class blog code java http
原文地址:http://www.cnblogs.com/zuoyuan/p/3779688.html