The number of reversing twice in leetcode simple question

subject

reverse An integer means inverting all its bits .

for example , reverse 2021 obtain 1202 . reverse 12300 obtain 321 , Do not retain leading zeros .
Give you an integer num , reverse num obtain reversed1 , Then reverse reversed1 obtain reversed2 . If reversed2 be equal to num , return true ; otherwise , return false .
Example 1：
Input ：num = 526
Output ：true
explain ： reverse num obtain 625 , Then reverse 625 obtain 526 , be equal to num .
Example 2：
Input ：num = 1800
Output ：false
explain ： reverse num obtain 81 , Then reverse 81 obtain 18 , It's not equal to num .
Example 3：
Input ：num = 0
Output ：true
explain ： reverse num obtain 0 , Then reverse 0 obtain 0 , be equal to num .
Tips ：
0 <= num <= 10^6
source ： Power button （LeetCode）

Their thinking

   As long as a bit is 0 And not a single digit , Then the reversal will not succeed , Otherwise, success .

class Solution:
    def isSameAfterReversals(self, num: int) -> bool:
        return False if str(num)[-1]=='0' and len(str(num))>1 else True