leedcode 二进制求和

自己写的

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        a_len=len(a)
        b_len=len(b)
        if a_len<b_len:
            temp=b
            b=a
            a=temp
        a_len = len(a)
        b_len = len(b)

        b_list=list()

        com_len=min(a_len,b_len)
        res_list = [0]*com_len
        res_fro_list=[0]*(a_len-com_len)

        a_fro = a[0:a_len-com_len]
        a_back = a[a_len-com_len:a_len]

        a_fro_list=list()
        a_fro_len = len(a_fro)

        a_back_list=list()
        a_back_len=len(a_back)

        for i in range(a_fro_len):
            a_fro_list.append(int(a_fro[i]))
        for i in range(a_back_len):
            a_back_list.append(int(a_back[i]))
        for i in range(b_len):
            b_list.append(int(b[i]))

        carry=0
        for i in range(com_len-1,-1,-1):
            res_list[i]=a_back_list[i]+b_list[i]+carry
            carry=res_list[i]//2
            res_list[i]=res_list[i]%2
        

        for i in range(a_len-com_len-1,-1,-1):
            res_fro_list[i]=a_fro_list[i]+carry
            carry=res_fro_list[i]//2
            res_fro_list[i]=res_fro_list[i]%2
        

        final_list=res_fro_list+res_list
        

        if carry:
            final_list.insert(0,carry)
        

        final_str=str()
        for i in range(len(final_list)):
            final_str+=str(final_list[i])


        return final_str



a = "1"
b = "111"
sol=Solution().addBinary(a,b)
print(sol)

gpt写的

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        a_len, b_len = len(a), len(b)

        # 使得 a 的长度不小于 b，可以避免后续的交换操作
        if a_len < b_len:
            a, b = b, a
            a_len, b_len = b_len, a_len

        res_list = []
        carry = 0

        # 从右向左遍历两个字符串
        for i in range(1, a_len + 1):
            bit_a = int(a[-i])
            bit_b = int(b[-i]) if i <= b_len else 0

            # 计算当前位的和
            current_sum = bit_a + bit_b + carry

            # 更新进位和当前位的值
            carry = current_sum // 2
            digit = current_sum % 2
            res_list.insert(0, str(digit))

        # 如果最终还有进位，插入到列表的开头
        if carry:
            res_list.insert(0, str(carry))

        return ''.join(res_list)