# Given a 32-bit signed integer, reverse digits of an integer. # Example 1: # Input: 123 # Output: 321 # Example 2: # Input: -123 # Output: -321 # Example 3: # Input: 120 # Output: 21 # Note: # Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows. # A very bad solution class Solution: def reverse(self, x: int) -> int: if x >= 2**31-1 or x <= -2**31: return 0 str_int = str(x) new_str = [] if str_int[0] == '-': new_str.append(str_int[0]) start = 1 else: start = 0 i = len(str_int) -1 str1 = "" while (i >= start): new_str.append(str_int[i]) i = i-1 reverse = int(str1.join(new_str)) if reverse >= 2**31-1 or reverse <= -2**31: return 0 return int(str1.join(new_str)) # Solution 2 # Time Complexity: O(log(x)) # Space Complexity O(1) INT_MAX = 2**31-1 INT_MIN = -2**31 class Solution: def reverse(self, x: int) -> int: if x >= INT_MAX or x <= INT_MIN: return 0 if x == 0: return x if x < 0: mult = -1 x = x*(-1) else: mult = 1 rev = 0 num = x while(num != 0 ): dig = num % 10 num = num // 10 if rev > INT_MAX/10: return 0 else: rev = rev*10 + dig return rev*mult