13. 罗马数字转整数

https://leetcode-cn.com/problems/roman-to-integer/

解法一：左加右减

这个思路很6，根据罗马数字的左加右减法则，比较s[i]和s[i+1]所表示的数字大小，若s[i]<s[i+1]，则res减去s[i]，否则res加上s[i]

class Solution:
    def romanToInt(self, s: str) -> int:
        #符号映射到数字
        dict = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        res = 0
        for i in range(len(s)):
            if i+1 < len(s) and dict[s[i]] < dict[s[i+1]]:
                res -= dict[s[i]]
            else:
                res += dict[s[i]]
        return res

先将s转换成相应数字，再比较，省去了查dict的时间，速度更快

class Solution:
    def romanToInt(self, s: str) -> int:
        dict = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        nums = [0] * len(s)
        for i in range(len(s)):    #转换成数字
            nums[i] = dict[s[i]]
        res = 0
        for i in range(len(s)):
            if i+1 < len(s) and nums[i] < nums[i+1]:
                res -= nums[i]
            else:
                res += nums[i]
        return res