9. Palindrome Number(判断整型数字是否是回文，直接暴力即可)

 1 class Solution:
 2     def isPalindrome(self, x):
 3         """
 4         :type x: int
 5         :rtype: bool
 6         """
 7         x =str(x)
 8 
 9         if(len(x)%2!=0):#数
10             for i in range(int(len(x)/2)+1):
11                 if(x[i]!=x[len(x)-i-1]):
12                     return False
13         else:
14             for i in range(int(len(x)/2)+1):
15                 if(x[i]!=x[len(x)-i-1]):
16                     return False
17         return True
18                     
19