问题: 5078. 负二进制数相加 给出基数为 -2 的两个数 arr1 和 arr2,返回两数相加的结果. 数字以 数组形式 给出:数组由若干 0 和 1 组成,按最高有效位到最低有效位的顺序排列.例如,arr = [1,1,0,1] 表示数字 (-2)^3 + (-2)^2 + (-2)^0 = -3.数组形式 的数字也同样不含前导零:以 arr 为例,这意味着要么 arr == [0],要么 arr[0] == 1. 返回相同表示形式的 arr1 和 arr2 相加的结果.两数的表示形式为…

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Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 二进制数想加,并且保存在string中,要注意的是如何将string和int之间互相转换,并且每位相加时,会有进位的可能,会影响之后相加的结果.而且两个输入string的长度也可能会不同.这时我们需要新建一个string,它的长度是两…

class Solution { public: string addBinary(string a, string b) { if(a==""&&b=="") return ""; if(a=="") return b; if(b=="") return a; ],*pb=&b[]; ,nb=; ; ,m=; while(*pa!='\0'){ pa++; na++; } pa--…

Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1 or 0. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010&q…

Given two binary strings, return their sum (also a binary string). Have you met this question in a real interview? Yes Example a = 11 b = 1 Return 100 LeetCode上的原题,请参见我之前的博客Add Binary. class Solution { public: /** * @param a a number * @param b a num…

def add_binary_nums(x,y): max_len = max(len(x), len(y)) x = x.zfill(max_len) y = y.zfill(max_len) result = '' carry = , -, -): r = carry r += r += result = ( == ') + result carry = : result = ' + result return result.zfill(max_len) print(add_binary_n…

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.…

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.…

1. 校验和 ICMP,IP,UDP,TCP报头部分都有checksum(检验和)字段.IP 首部里的校验和只校验首部:ICMP.IGMP.TCP和UDP首部中的校验和校验首部和数据. UDP和TCP的校验和不仅要对整个IP协议负载(包括UDP/TCP协议头和UDP/TCP协议负载)进行计算,还要先对一个伪协议头进行计算:先要填充伪首部各个字段,然后再将UDP/TCP报头及之后的数据附加到伪首部的后面,再对伪首部使用校验和计算,所得到的值才是UDP/TCP报头部分的校验和. 伪首部结构如下: 0…