Given two binary strings, return their sum (also a binary string).The input strings are both non-empty and contains only characters 1 or 0.
Example 1:
Input: a = "11", b = "1"
Output: "100"
Example 2:
Input: a = "1010", b = "1011"
Output: "10101"
思路
这道题和字符串加法思路是一样的，两个字符串都从尾向前遍历，使用溢出标志量进行记录是否存在溢出。直到遍历完毕。时间复杂度为O(n+m), n,m为a，b字符串的长度，空间复杂度为O(N), N为n，m中最大值加1。
解题思路
1 class Solution(object):
2 def addBinary(self, a, b):
3 """
4 :type a: str
5 :type b: str
6 :rtype: str
7 """
8 res, flow= '', 0 # 设置结果变量和溢出变量
9 i, j = len(a) - 1, len(b) - 1 # 两个字符串长度
10 while i >= 0 or j >= 0 or flow: # 循环变量
11 curval = (i >= 0 and a[i] == '1') + (j >= 0 and b[j] == '1') # 每一位相加的结果
12 flow, rem = divmod(curval + flow, 2) # 存储结果
13 res = `rem` + res
14 i -= 1
15 j -= 1
16 return res