[Leetcode][python3] 7. Reverse Integer

Problem:

 

Code 1 :

I used String API in Python.

First, convert Integer X to String

Second, check if last character is 0 or not reverse string 

class Solution:
    def reverse(self, x: int) -> int:
        
        if x == 0:
            return 0
        
        s = str(x)
        negativeFlag = 1
        if s[0] == '-':
            negativeFlag = -1
            s = s[1:]  
                   
        index = 0
        while index<len(s) and s[index] =='0':
            index += 1
        
        s = s[index:]
        result = negativeFlag*int(s[::-1])
        if result > (2**31-1) or (-2**31)>result :
            return 0
        return result

 

 

Code 2: 

I'm not sure if it is right way that interviewer wants for me to do ...

So I tried to solve it in a different way

In the second code below, I used divide operation. (Integer division)

 

Divided X by 10 

 The remainder number would be new number in the result.

class Solution:
    def reverse(self, x: int) -> int:
        # check negative or positive
        sign = 1 if x>0 else -1
            
        x= abs(x)
        
        # reverse
        tail = 0
        result = 0
        while x: 
            x, mod = divmod(x, 10)
            result = result*10 + mod            
            #print(f'x:{x} mod:{mod} result:{result}')
            
        result = sign*result
        if result < (-2**31) or result > (2**31-1):
            return 0
        else:
            return result

 

 

English Vocabulary Expression

modulus operator : 나머지 구하는 것 %

% : percent sign

remainder : 나머지

11/4 = 2...3

"Eleven divided by four is two remainder three"