013.Roman to Integer.md

13.Roman to Integer

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

难度: Easy

思路：

integer to Roman 是 Medium，这个roman to integer是easy

所以用的傻方法，先特殊处理4，9，40，90，400，900，再加上剩下的数字

AC代码

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        lookup1 = {'CM':900,'CD':400,'XC':90,'XL':40,'IX':9,'IV':4}
        lookup2 = {'M':1000,  'D':500,  'C':100,  'L':50,  'X':10, 'V':5,  'I':1}

        num = 0
        for i in lookup1.keys():
        	if i in s:
        		num += lookup1[i]
        		s = s.replace(i,'')

        for char in s:
        	num += lookup2[char]

        return num

然后可以有更好的方法：

从前往后扫描，用一个临时变量记录分段数字。

如果当前比前一个大，说明这一段的值应当是这个值减去上一个值。比如IV = 5-1 =4; 否则，将当前值加入到结果中，然后开始下一段记录，比如VI = 5 + 1, II = 1 +1

所以这也就是罗马数字的基础，感觉？这样才不会读串？

AC代码

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        lookup = {'M':1000,  'D':500,  'C':100,  'L':50,  'X':10, 'V':5,  'I':1}

        num = 0

        for i in range(len(s)):
        	if i > 0 and lookup[s[i]] > lookup[s[i-1]]:
        		num += lookup[s[i]] - 2 * lookup[s[i-1]]
        	else:
        		num += lookup[s[i]]

        return num