https://leetcode.com/problems/reverse-integer/
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Update (2014-11-10):
Test cases had been added to test the overflow behavior.
Code
import re
class Solution(object):
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
m = re.match('(^[\+\-]*)(\d+)', str(x))
sign = 1
if m and m.group(1):
sign = -(ord(m.group(1)) - 44) # '-' ascii: 45 '+' ascii: 43
if m and m.group(2):
num = int(m.group(2)[::-1])
num = num * sign
if num > 2147483647 or num < -2147483648:
return 0
return num
return 0
Idea
Simply using regex to find the two parts in the original string: sign and digits. I am a little cheating by calling `int()` directly: it automatically handles the situations when `m.group(2)[::-1]` starts with multiple zeros.
Spoilers for the problem are really important. This is an easy problem but with some corner cases you need to consider.