[Leet Code] 9. Palindrome Number - Python

https://leetcode.com/problems/palindrome-number/description/

Level : Easy
Topic : Math

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # 음수라면 False
        # 0이라면 true
        # 양수라면
        #   1. 한자리수 -> true
        #   2. not 한자리수
        #      n = len(str(x))//2 -> 반복문을 n번 돌며 앞뒤가 다르면 false 같으면 반복문 계속돌기
        if x < 0 : return False
        #elif x == 0 : return True
        #elif len(str(x)) == 1: return True
        else:
            x = str(x)
            n = len(x) // 2
            j = len(x)-1
            for i in range(n):
                if x[i] != x[j]: return False
                j -= 1
            return True
            
    		# Time Complexity : O(n/2) = O(n)
            # Space Complextiy : O(1)