leetcode67：二进制求和

最新推荐文章于 2022-09-22 13:28:46 发布

原创最新推荐文章于 2022-09-22 13:28:46 发布 · 128 阅读

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本文介绍了一种通过补零使两个二进制字符串长度相等，然后逐位相加并考虑进位的算法实现。该算法适用于字符串形式的二进制数加法运算。

思想：

将通过补零的方式使得字符串a和字符串b长度一致，然后从字符串a和字符串b最后一位和进位carry进行加，然后判断属于哪种情况，最后注意如果carry为1需要加入

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        carry = 0
        c = []
        a = a.zfill(max(len(a), len(b)))
        b = b.zfill(max(len(a), len(b)))
        for i in range(-1, -(len(a)+1), -1):
            if int(a[i])+int(b[i])+carry == 2:
                c.insert(i, '0')
                carry = 1
            elif int(a[i])+int(b[i])+carry == 3:
                c.insert(i, '1')
                carry = 1
            elif int(a[i])+int(b[i])+carry == 1:
                c.insert(i, '1')
                carry = 0
            else:
                c.insert(i, '0')
                carry = 0
        if carry == 1:
            c.insert(0, '1')
        return "".join(c)


if __name__ == "__main__":
    print(Solution().addBinary("100", "110010"))

小菜鸟的思想还是对的，只是用的方法有点浪费时间，接下来仿照大佬的编程习惯进行改进

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        carry = 0
        c = ''
        a = a.zfill(max(len(a), len(b)))
        b = b.zfill(max(len(a), len(b)))
        for i in range(-1, -(len(a)+1), -1):
            if int(a[i])+int(b[i])+carry == 2:
                c += '0'
                carry = 1
            elif int(a[i])+int(b[i])+carry == 3:
                c += '1'
                carry = 1
            elif int(a[i])+int(b[i])+carry == 1:
                c += '1'
                carry = 0
            else:
                c += '0'
                carry = 0
        if carry == 1:
            c += '1'
        return c[::-1]

似乎快了一些些