Leetcode brush question 𞓜 13 Roman numeral to integer

A problem I did three months ago , It has become today's daily question , It seems that if you don't record it, you will soon forget .

Title Description

Address ： Roman numeral to integer

Solution 1

Although the title gives six special cases of Roman numerals to integers , But actually , The core of this question is , In the process of converting Roman array to integer , You need to observe the size relationship between the two numbers before and after , If the previous number is smaller than the following number , It takes so long to subtract the number in front of this , conversely , Just add this number . The logic is simple , Then let's correspond the numbers represented by each letter , Then you can traverse the given string .

class Solution:
    def romanToInt(self, s: str) -> int:
        prenum = self.getChar(s[0])
        n = len(s)
        num = 0
        for i in range(1,n):
            t = self.getChar(s[i])
            if t > prenum:
                num = num - prenum
            else:
                num = num + prenum
            prenum = t
        num += prenum
        return num


    def getChar(self, c):
        if c == "I":
            return 1
        elif c == "V":
            return 5
        elif c == "X":
            return 10
        elif c == "L":
            return 50
        elif c == "C":
            return 100
        elif c == "D":
            return 500
        elif c == "M":
            return 1000
        else:
            return 0

The correspondence between letters and numbers can also be directly used in the dictionary ：

class Solution(object):
    def romanToInt(self,s):
        """ :type s: str :rtype: int """
        num_dict = {
    'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
        sum = 0
        prenum = num_dict[s[0]]
        for i in range(1,len(s)):
            num = num_dict[s[i]]
            if prenum<num:
                sum = sum - prenum
            else:
                sum = sum + prenum
            prenum = num
        return sum + prenum