old_contents: Determine whether an integer is a palindrome. Do this without extra space.
  click to show spoilers.
  Some hints:
  Could negative integers be palindromes? (ie, -1)
  If you are thinking of converting the integer to string, note the restriction of using extra space.
  You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", 
  you know that the reversed integer might overflow. How would you handle such case?
  There is a more generic way of solving this problem.
  
  class Solution:
      # @return a boolean
      def isPalindrome(self, x):
          if x < 0:
              return False
          div = 10
          while x > div:
              div *= 10 
          div /= 10 
          while x > 0:
              if x / div != x % 10:  # compare the first number and last number
                  return False
              x = (x % div )/10
              div /= 100
          return True
  
new_contents: Determine whether an integer is a palindrome. Do this without extra space.
  click to show spoilers.
  Some hints:
  Could negative integers be palindromes? (ie, -1)
  If you are thinking of converting the integer to string, note the restriction of using extra space.
  You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", 
  you know that the reversed integer might overflow. How would you handle such case?
  There is a more generic way of solving this problem.
  
  # Best answer ! 
  class Solution:
      # @return a boolean
      def isPalindrome(self, x):
          if x < 0 or (x != 0 and x % 10 == 0): return False # except 10, 100
          sum  = 0
          while x > sum:
              sum = sum * 10 + x % 10
              x = x / 10
          return (x == sum) or (x == sum/10)
  
  class Solution:
      # @return a boolean
      def isPalindrome(self, x):
          if x < 0:
              return False
          div = 10
          while x > div:
              div *= 10 
          div /= 10 
          while x > 0:
              if x / div != x % 10:  # compare the first number and last number
                  return False
              x = (x % div )/10
              div /= 100
          return True
  
subject: Update Palindrome_Number.py