[LeetCode][python3]0013. Roman to Integer

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N2I -2020.03.18

1. My first solution

class Solution:
    def romanToInt(self, s: str) -> int:
        rom=["M","D","C","L","X","V","I"]
        digit=[1000,500,100,50,10,5,1]
        result=0
        
        if s.find("CM")>=0 or s.find("CD")>=0:
            result-=200
        if s.find("XC")>=0 or s.find("XL")>=0:
            result-=20
        if s.find("IX")>=0 or s.find("IV")>=0:
            result-=2
        for c in s:
            for r,d in zip(rom,digit):
                if c==r:
                    result+=d
        return result

N2I -2020.03.18

Explanation:

Using find() function to solve the problem. When C+M=1000 and CM=900, this makes the result have a 200 gap. So we minus it first then count it as the same way. Same as CD,XC,XL,IX,IV.

2. Other solution

class Solution:
    def romanToInt(self, s: str):
        symbol={'I':1,'V':5,'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        output=[0]
        for i in s:
            if symbol[i]>output[-1]:
                output[-1]=symbol[i]-output[-1]
            else:
                output.append(symbol[i])
        return sum(output)

N2I -2020.03.18

Explanation:

The solution is using dict to save the pattern, and check out the special cases when smaller symbol is in front of a bigger symbol.