class Solution { public: string addBinary(string a, string b) { int car = 0; int cur_a = a.length()-1; int cur_b = b.length()-1; int num =0; string ret; int toa = 0; int tob = 0; while(cur_a>=0 && cur_b>=0) { if(cur_a >= 0) { toa = char2int(a[cur_...
publicclassSolution {publicString addBinary(String a, String b) {if(a.equals("") && b.equals(""))return"0";if(a ==null|| a.equals("") || a.equals("0"))returnb;if(b ==null|| b.equals("") || b.equals("0"))returna;char[] a1 =a.toCharArray();char[] b1 =b.toCharArr...
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...
Add Binary Desicription Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". Solution 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public: string addBinary(string a, string b) { string res; int a_in...
class Solution: def addBinary(self, a: str, b: str) -> str: sumInt = int(a, 2) + int(b, 2) sumBin = bin(sumInt) #string starts with '0b' return sumBin[2:] # equally, but more precise # return bin( int(a, 2) + int(b, ) )[2:] # return '{:b}'.format(int(a...
class Solution(object): def addBinary(self, a, b): """ :type a: str :type b: str :rtype: str """ result = [] val = 0 carry = 0 lenA = len(a) lenB = len(b) if lenA < lenB: # 注意坑:确保A比B长, 所以值长度也要交换 a, b = b, a lenA, lenB = lenB, lenA for...
leetcode -- Add Binary -- 简单要了解 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...
public class Solution { public String addBinary(String a, String b) { int i = a.length() - 1, j = b.length() - 1, carry = 0; StringBuilder sb = new StringBuilder(); while(i >=0 || j >=0){ int m = i >= 0 ? a.charAt(i) - '0' : 0; ...
class Solution { public String addBinary(String a, String b) { StringBuilder res = new StringBuilder(); // 返回结果 int i = a.length() - 1; // 标记遍历到 a 的位置 int j = b.length() - 1; // 标记遍历到 b 的位置 int carry = 0; // 进位 ...
FindHeaderBarSize FindTabBarSize FindBorderBarSize Given two binary stringsaandb, returntheir sum as a binary string. Example 1: Input:a = "11", b = "1"Output:"100" Example 2: Input:a = "1010", b = "1011"Output:"10101" Constraints:...