題目說明:給你一個數字，要你判斷此數字是否為回文

Case 1:
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Case 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Case 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

解題思路:最簡單的方式就是將數字轉換成字串並且將字串反轉比對即可。另一種方式就是使用迴圈，從個位數字乘上x的位數減一，十位數字乘上x的位數減二，以此類推...

附上程式碼和註解
Java

class Solution {
    public boolean isPalindrome(int x) {
        StringBuilder builder=new StringBuilder();//建立stringbuilder物件
        String y=String.valueOf(x);//數字轉字串
        builder.append(y);//加入字串
        builder.reverse();//反轉
        if(y.equals(builder.toString())&&x>=0){
            return true;
        }
        return false;
    }
}

Python

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x<0:
            return False＃只要x<0一定是錯的(因為反轉會多一個負號)
        else:
            n=x
            digit=len(str(x))-1
            s=0
            while x>0:
                s+=pow(10,digit)*(x%10)
                x//=10
                digit-=1
            return s==n