""" 假设我们的环境只能存储得下 32 位的有符号整数，则其数值范围为 [−2^31, 2^31 − 1]。请根据这个假设，如果反转后整数溢出那么就返回 0。 """ class Solution: def reverse(self, x): """ input: x Int output: Int """ res = 0 sign = -1 if x < 0 else 1 boundary = 1<<31 if x < 0 else (1<<31) - 1 num = sign * x while num > 0: d = num % 10 res = res*10 + d if res > boundary: return 0 num = num // 10 return sign*res if __name__ == '__main__': sol = Solution() num = 1463847412 res = sol.reverse(num) print(res)