Leetcode67: Add Binary
具体题目如下：
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1：先算出a,b求和的十进制数，再将十进制转换为二进制即可，思路简单，代码也比较简洁：
class Solution:
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
s = 0
for i in range(len(a)):
if a[i] == '1':
s += 2**(len(a)-i-1)
for i in range(len(b)):
if b[i] == '1':
s += 2**(len(b)-i-1)
c = ''
if s==0:
return '0'
while(s): ##这几行就是把十进制转换为二进制，若想改为其他进制，则把余数改一下就可以了；
d = s%2
s = s//2
c += str(d)
return c[::-1]
思路2：一开始我对a,b两个字符串的每个字符从低位到高位分别求和，逢2进1，可以AC，但是判断边界条件比较麻烦，具体看代码：
class Solution:
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
a = a[::-1]
b = b[::-1]
length = min(len(a), len(b))
c = ''
t = 0
for i in range(length):
if int(a[i]) + int(b[i]) + t >= 2:
c += str(int(a[i]) + int(b[i]) + t-2)
t = 1
else:
c += str(int(a[i]) + int(b[i]) + t)
t = 0
i = 0
if len(a) > len(b):
while (length + i < len(a) and t + int(a[length + i]) >= 2):
c += str(t + int(a[length + i]) - 2)
t = 1
i += 1
if length + i == len(a):
if t ==1:
c += '1'
else:
c += str(t + int(a[length + i]))
if length+i+1<len(a):
c += a[length + i+1:]
else:
while (length + i < len(b) and t + int(b[length + i]) >= 2):
c += str(t + int(b[length + i]) - 2)
t = 1
i += 1
if length + i == len(b):
if t ==1:
c += '1'
else:
c += str(t + int(b[length + i]))
if length + i + 1 < len(b):
c += b[length + i + 1:]
return c[::-1]