将一个整数中的数字进行颠倒,当颠倒后整数溢出时,返回 0
参考代码:
class Solution: def reverseInteger(self, n): if n == 0: return 0 neg = 1 if n < 0: neg, n = -1, -n revers = 0 while n > 0: reverse = reverse * 10 + n % 10 n = n // 10 reverse = reverse * neg if reverse < - (1 << 31) or reverse > (1 << 31) - 1: return 0 return reverse