题目如下：
解题思路：首先将较短的输入前面补0，保证a和b长度一样，然后逐位相加再加上进位的值。如果和为3，当前位值为1，进位1；如果为2，当前位值为0，进位为1；否则不进位，当前位值即为和的值。
代码如下：
class Solution(object): def addBinary(self, a, b): """ :type a: str :type b: str :rtype: str """ diff = abs(len(a) - len(b)) if len(a) < len(b): a = ‘0‘*diff + a elif len(b) < len(a): b= ‘0‘*diff + b carry = 0 res = ‘‘ for i,j in zip(a[::-1],b[::-1]): v = int(i) + int(j) + carry if v == 3: v = 1 carry = 1 elif v == 2: v = 0 carry = 1 else: carry = 0 res = str(v) + res if carry > 0: res = str(carry) + res return res
原文：https://www.cnblogs.com/seyjs/p/9275618.html