簡單題，如果不轉string要犧牲一些時間復雜度：

Submission Detail

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # solve it without converting the integer to a string
 

        # deal 0
        if x==None:
            return False

        if x < 0:
            return False
        if x == 0:
            return True
        # noraml
        order = []
        while (True):
            remain = x % 10
            div = x // 10
            if div == 0 and remain ==0:
                
 
break
            else:
                order.append(remain)
                x = x // 10

        # is Palindrome
        # notice lengthS >=
        lengthS = len(order)
        if lengthS % 2 == 1:
            # odd
            list1 = order[0:lengthS // 2 + 1]
            list2 = order[lengthS // 2:]
            list2.reverse()
            
 
for index in range(len(list1)):
                if list1[index] == list2[index]:
                    if index == len(list1) -1:
                        return True
                else:
                    # no palindrome
                    return False
        else:
            # even
            list1 = order[0:lengthS // 2]
            list2 = order[lengthS // 2:]
            list2.reverse()
            for index in range(len(list1)):
                if list1[index] == list2[index]:
                    if index == len(list1)-1:
                        return True
                else:
                    # no palindrome
                    return False

[leetcode]9. Palindrome Number