old_contents: # Time: O(logn)
  # Space: O(1)
  #
  # Reverse digits of an integer.
  # 
  # Example1: x = 123, return 321
  # Example2: x = -123, return -321
  # 
  # click to show spoilers.
  # 
  # Have you thought about this?
  # Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
  # 
  # If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
  # 
  # Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, 
  # then the reverse of 1000000003 overflows. How should you handle such cases?
  # 
  # Throw an exception? Good, but what if throwing an exception is not an option? 
  # You would then have to re-design the function (ie, add an extra parameter).
  #
  
  class Solution:
      # @return an integer
      def reverse(self, x):
          ans = 0
          if x >= 0:
              while x:
                  ans = ans * 10 + x % 10
                  x /= 10
              return ans
          else:
              return -self.reverse(-x)
          
  if __name__ == "__main__":
      print Solution().reverse(123)
      print Solution().reverse(-321)
new_contents: # Time: O(logn)
  # Space: O(1)
  #
  # Reverse digits of an integer.
  # 
  # Example1: x = 123, return 321
  # Example2: x = -123, return -321
  # 
  # click to show spoilers.
  # 
  # Have you thought about this?
  # Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
  # 
  # If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
  # 
  # Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, 
  # then the reverse of 1000000003 overflows. How should you handle such cases?
  # 
  # Throw an exception? Good, but what if throwing an exception is not an option? 
  # You would then have to re-design the function (ie, add an extra parameter).
  #
  
  class Solution:
      # @return an integer
      def reverse(self, x):
          ans = 0
          if x >= 0:
              while x:
                  ans = ans * 10 + x % 10
                  x /= 10
              return ans if ans <= 2147483647 else 0  # Handle overflow.
          else:
              return -self.reverse(-x)
          
  if __name__ == "__main__":
      print Solution().reverse(123)
      print Solution().reverse(-321)
  
subject: Update reverse-integer.py