I'm working with lists and I'm trying to access the next value in an if statement like this: if((lst[i+1] != 'V') or (lst[i+1] != 'X')):. But as you can see, this is super prone to errors. And it did cause an IndexError.
What is the best way to call the next item in a list without trying to breach the list's size?
Basically, its a LeetCode assignment asking to convert Roman numbers to regular numbers. This is my full code:
class Solution:
def romanToInt(self, s: str) -> int:
lst = list(s)
counter = 0
for i in range(len(lst)):
if(lst[i] == 'I'):
if((lst[i+1] != 'V') or (lst[i+1] != 'X')):
counter += 1
elif(lst[i+1] == 'V'):
counter += abs(1-5)
elif(lst[i+1] == 'X'):
counter += abs(1-10)
if(lst[i] == 'V'):
if(lst[i-1] != 'I'):
counter += 5
else:
counter += abs(1-5)
if(lst[i] == 'X'):
if((lst[i+1] != 'L') or (lst[i+1] != 'C')):
counter += 10
elif(lst[i+1] == 'L'):
counter += abs(10-50)
elif(lst[i+1] == 'C'):
counter += abs(10-100)
if(lst[i] == 'L'):
if(lst[i-1] != 'X'):
counter += 50
else:
counter += abs(10-50)
if(lst[i] == 'C'):
if((lst[i+1] != 'D') or (lst[i+1] != 'M')):
counter += 100
elif(lst[i+1] == 'D'):
counter += abs(100-500)
elif(lst[i+1] == 'M'):
counter += abs(100-1000)
if(lst[i] == 'D'):
if(lst[i-1] != 'C'):
counter += 500
else:
counter += abs(100-500)
if(lst[i] == 'M'):
if(lst[i-1] != 'C'):
counter += 1000
else:
counter += abs(100-1000)
return counter
I don't want any deep help. I just want to know how to come across this one issue as I couldn't find anything relative to conditional statements.
In you code you need to replace your for loop:
for i in range(len(lst))
with
for i in range(len(lst))
to avoid going out of range since you are accessing the lst lst at both i+1, and i-1 in your loop.
For another example of converting Roman numerals to decimals check out: https://www.geeksforgeeks.org/converting-roman-numerals-decimal-lying-1-3999/