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; ...
参考:https://leetcode.com/problems/add-binary/discuss/24475/Short-code-by-c%2B%2B classSolution{public:stringaddBinary(stringa,stringb){stringresultStr ="";intcarry =0, aIndex = a.size() -1, bIndex = b.size() -1;//初始化余数和a,b索引值while(aIndex >=0|| bIndex >=0|| carr...
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 """ if not a: return b elif not b: return a tmp_...
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) { c += i >= 0 ? a[i--] - '0' : 0; ...
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) { ...
🔥 Join LeetCode to Code! View your Submission records hereRegister or Sign In Ln 1, Col 1 You need to Login / Sign up to run or submit Case 1Case 2 a = "11" b = "1" 1 2 3 4 "11" "1" "1010" "1011" Source