Can you solve this real interview question? Add Binary - Given two binary strings a and b, return their sum as a binary string. Example 1: Input: a = "11", b = "1" Output: "100" Example 2: Input: a = "1010", b = "1011" Output: "10101" Constr
#include<iostream>#include<string>usingnamespacestd;classSolution{public:stringAddBinary(string a, string b){size_tsize = a.size() > b.size() ? a.size() : b.size();reverse(a.begin(), a.end());reverse(b.begin(), b.end());intCarryBit =0;// 进位string result;// 用于存放结果...
Given two binary strings, return their sum (also a binary string). For example, a ="11" b ="1" Return"100". 解题思路1:判断当前字符所表示的数字,产生输出和进位。缺点:程序比较复杂。 代码1: class Solution { public: string addBinary(string a, string b) { char carry='0'; string c; ...
usestd::iter; implSolution{ pubfnadd_binary(a:String,b:String)->String{ //进位,初始为0 letmutcarry=0; //收集a和b按位加的结果 letmutresult=a //返回底层的字节切片 .as_bytes() //转换成迭代器 .iter() //提前反向迭代(后面会加上无限的'0',所以不能后面同时反向) .rev() //在a后面...
https://leetcode.com/problems/add-binary/ 知道二进制加法原理即可,这里只需要知道进位是除数,余数是结果就行。最后不要忽略reg里面的值 class Solution(object): def addBinary(self, a, b): """ :type a: str :type b: str :rtype: str
AI检测代码解析 class Solution { public: string addBinary(string a, string b) { string s; s.reserve(a.size() + b.size()); int c = 0, i = a.size() - 1, j = b.size() - 1; while(i >= 0 || j >= 0 || c == 1) ...
classSolution:defaddBinary(self,a:str,b:str)->str:sumInt=int(a,2)+int(b,2)sumBin=bin(sumInt)#string starts with '0b'returnsumBin[2:]# equally, but more precise# return bin( int(a, 2) + int(b, ) )[2:]# return '{:b}'.format(int(a, 2) + int(b, 2))# return f"...
Add Binary 二进制数求和 128 -- 6:56 App [LeetCode] 27. Remove Element 移除元素 102 -- 15:04 App [LeetCode] 71. Simplify Path 简化路径 23 -- 9:03 App [LeetCode] 16. 3Sum Closest 最接近的三数之和 20 -- 13:44 App [LeetCode] 15. 3Sum 三数之和 26 -- 15:53 App ...
Leetcode - Add Binary Paste_Image.png My code: public class Solution { public String addBinary(String a, String b) { if (a == null || a.length() == 0 || b == null || b.length() == 0) return null; int len = Math.max(a.length(), b.length()) + 1;...
Add Binary https://leetcode.com/problems/add-binary/description/ 主要是全加器的公式: s = abc c = (a^b)&c | (a&b) 其他的没什么 就去一下零处理 以及颠倒一下列表 classSolution(object):defaddBinary(self,a,b):""" :type a: str...