reverse-integer.py

# Time: O(logn) = O(1)
# 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(object):
 def reverse(self, x):
 """
 :type x: int
 :rtype: int
 """
 if x < 0:
 return -self.reverse(-x)

 result = 0
 while x:
 result = result * 10 + x % 10
 x //= 10
 return result if result <= 0x7fffffff else 0 # Handle overflow.

 def reverse2(self, x):
 """
 :type x: int
 :rtype: int
 """
 if x < 0:
 x = int(str(x)[::-1][-1] + str(x)[::-1][:-1])
 else:
 x = int(str(x)[::-1])
 x = 0 if abs(x) > 0x7FFFFFFF else x
 return x

 def reverse3(self, x):
 """
 :type x: int
 :rtype: int
 """
 s = cmp(x, 0)
 r = int(`s * x`[::-1])
 return s * r * (r < 2 ** 31)


if __name__ == "__main__":
 print Solution().reverse(123)
 print Solution().reverse(-321)