🟢 剑指 Offer II 002. 二进制加法

LeetCode 提示

题目难度 简单

原题链接 🔗 leetcode

解析 1.md

庸俗的解法

题解 1.py

from collections import deque
class Solution:    def addBinary(self, a: str, b: str) -> str:        res = deque('')
        idxa, idxb = len(a)-1, len(b)-1        c = 0        while idxa >= 0 and idxb >= 0:            pos = int(a[idxa]) + int(b[idxb]) + c            c = pos // 2            res.appendleft(str(pos % 2))            idxa, idxb = idxa-1, idxb-1                while idxa >= 0:            pos = int(a[idxa]) + c            c = pos // 2            res.appendleft(str(pos % 2))            idxa -= 1                while idxb >= 0:            pos = int(b[idxb]) + c            c = pos // 2            res.appendleft(str(pos % 2))            idxb -= 1                if c:            res.appendleft(str(c))                return ''.join(list(res))

解析 2.md

python有自带转二进制字符串的方法，再加上位运算算加法即可

# 自带方法介绍bin(2) # '0b10'
int('10', 2) # 按二进制转化 得到2

题解 2.位运算.py

class Solution:    def addBinary(self, a: str, b: str) -> str:        x, y = int(a, 2), int(b, 2)
        while y:            res = x ^ y            carry = (x & y) << 1            x, y = res, carry                return bin(x)[2:]