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; ...
直接用二进制相加 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[...
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...
class Solution { public: string addBinary(string a, string b) { int temp,len_a,len_b; len_a = a.size(); len_b = b.size(); string::iterator itr_a = a.end() - 1; string::iterator itr_b = b.end() - 1; string str_a =a; string str_b =b; if(len_a<len_b) { a ...
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
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...
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* int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // Start typing your C/C++ solution below if(l1 == NULL && l2 == NULL) { ...
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: 1 <= a.length, b.length <= 104