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【leetcode 简单】第十六题 二进制求和

程序员文章站 2022-05-27 08:08:29
给定两个二进制字符串,返回他们的和(用二进制表示)。 输入为非空字符串且只包含数字 1 和 0。 示例 1: 输入: a = "11", b = "1" 输出: "100" 示例 2: 输入: a = "1010", b = "1011" 输出: "10101" class Solution: de ......

给定两个二进制字符串,返回他们的和(用二进制表示)。

输入为非空字符串且只包含数字 1 和 0

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        a_len=len(a)
        b_len=len(b)
        result=''
        next ='0'
        while a_len and b_len :
            if sum([int(next),int(a[a_len-1]),int(b[b_len-1])]) == 3:
                result +='1'
                next = '1'
            elif sum([int(next),int(a[a_len-1]),int(b[b_len-1])]) == 2:
                result+='0'
                next ='1'
            elif sum([int(next),int(a[a_len-1]),int(b[b_len-1])]) == 0:
                result+='0'
                next ='0'
            elif sum([int(next),int(a[a_len-1]),int(b[b_len-1])]) < 2:
                result+='1'
                next ='0'
            a_len-=1
            b_len-=1
        new = a[:a_len] if a_len > b_len else b[:b_len]
        if new:
            for i in new[::-1]:
                if sum([int(next),int(i)]) == 2:
                    result+='0'
                    next ='1'
                elif sum([int(next),int(i)]) == 1:
                    result+='1'
                    next ='0'
                else:
                    result+='0'
                    next ='0'
        if next != 0:
            result+=next
        return (result[::-1])
class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """

        return format(int(a,2)+int(b,2),"b")