1 问题

给你两个二进制字符串 a 和 b ，以二进制字符串的形式返回它们的和。

示例 1：

输入:a = “11”, b = “1”
输出：“100”

示例 2：

输入：a = “1010”, b = “1011”
输出：“10101”

2 答案

自己写的，先转成十进制，再求和，再转成二进制

class Solution:def addBinary(self, a: str, b: str) -> str:int_a, int_b = 0, 0c = ''for i in range(len(a)):int_a = int_a + int(a[i]) * 2 ** (len(a)-i-1)for j in range(len(b)):int_b = int_b + int(b[j]) * 2 ** (len(b)-j-1)int_c = int_a + int_bif int_c==0 : return "0" while int_c:c = str(int_c % 2) + cint_c //= 2return c

官方解

1. 一行写法

class Solution:def addBinary(self, a: str, b: str) -> str:return bin(int(a, 2) + int(b, 2))[2:]  # int(a, 2) 将二进制字符串转成整数# bin 将整数转换为二进制# 对结果如"0b100"进行切片，去掉前面的’0b‘

2. 二进制两数相加

class Solution:def addBinary(self, a: str, b: str) -> str:res, p = '', 0d = len(b) - len(a)a = '0'*d + a  # 让a和b长度相同，如果d小于零，则不会补0b = '0'*-d + b for i, j in zip(a[::-1],b[::-1]):  # 翻转切片，从后往前加s = int(i) + int(j) + p  # 二进制两数相加res = str(s%2) + resp = s // 2return str(p)+res if p else res

https://leetcode.cn/problems/add-binary/solutions/7221/python-1xing-nei-zhi-han-shu-fei-nei-zhi-jie-fa-by/

3. 也可以使用 divmod

class Solution:def addBinary(self, a: str, b: str) -> str:res, carry = '', 0i = len(a) - 1j = len(b) - 1while i>=0 or j>=0 or carry:tmp1 = int(a[i]) if i>=0 else 0 tmp2 = int(b[j]) if j>=0 else 0 carry, t = divmod(tmp1+tmp2+carry, 2) # carry除以2的商，t为余数res = str(t) + resi -= 1j -= 1return res

https://leetcode.cn/problems/add-binary/solutions/6200/mo-ni-guo-cheng-by-powcai-4/