題目說明：給一個數字並將其反轉兩次後是否等於原本的數字;反轉的定義:1234反轉變成4321，100變成1(將前面的0去除掉)

Case 1:
Input: num = 526
Output: true
Explanation: Reverse num to get 625, then reverse 625 to get 526, which equals num.

Case 2:
Input: num = 1800
Output: false
Explanation: Reverse num to get 81, then reverse 81 to get 18, which does not equal num.

Case 3:
Input: num = 0
Output: true
Explanation: Reverse num to get 0, then reverse 0 to get 0, which equals num.

解題的思路:其實從題目對於反轉的定義以及測資就可以發現只要尾數有0的數字，反轉以後其位數(100原本是三位數，反轉過後變成一位樹)一定會跟原本的不一樣，因此反轉第二次時得到的答案也會和原本不一樣，所以這題只是單純要你判斷這個數字除以10的餘數是否為0而已

附上程式碼
Java

class Solution {
    public boolean isSameAfterReversals(int num) {
        if (num==0) return true;
        return num%10!=0;
    }
}

Python

class Solution:
    def isSameAfterReversals(self, num: int) -> bool:
        if num==0:
            return True
        return not num%10==0