算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

python 代码

class Solution(object):
    def add(self, num1, num2):
        """
        :type num1: int
        :type num2: int
        :rtype: int
        """
        n1 = (num1 ^ num2) & 0xFFFFFFFF
        n2 = ((num1 & num2) << 1) & 0xFFFFFFFF
        n3 = n1 ^ n2
        if n2 == 0:
            return n3 if n3 <= 0x7fffffff else ~(n3^0xFFFFFFFF)
        else:
            return self.add(n1,n2)