Leetcode 67:Add Binary(二进制求和) (python.java) 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. 给定两个二进制字符串,返回他们的和(用二进制表示). 输入为非空字符串且只包含数字 1 和 0. Example 1: Input…

题目: 二进制求和 给定两个二进制字符串,返回他们的和(用二进制表示). 样例 a = 11 b = 1 返回 100 解题: 和求两个链表的和很类似 考虑进位,考虑最后一项的进位 0+0 = 0 不需要进位 0+1 = 1 不需要进位 1+1 =0  进位 1 同时注意 低位进1,高位时1+1的情况,直接加就是3了,这个需要进位1 ,原位的结果也是1的情况 Java程序: public class Solution { /** * @param a a number * @param b a…

［抄题］: 以字符串的形式给出两个非负整数 num1 和 num2,返回 num1和 num2 的和. 比如一个50位+一个100位. 给定 num1 = "123",num2 = "45" 返回 "168" [暴力解法]: 时间分析: 空间分析: ［思维问题］: ［一句话思路］: ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: sum求和或者循环退出的条件都是i j &g…

给定两个二进制字符串,返回他们的和(用二进制表示).案例:a = "11"b = "1"返回 "100" .详见:https://leetcode.com/problems/add-binary/description/ Java实现: class Solution { public String addBinary(String a, String b) { String s=""; int c=0; int i=a.len…

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…

Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 思路:二进制加法,比較简单.代码例如以下: public class Solution { public String addBinary(String a, String b) { int len = Math.max(a.l…

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&qu…

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). For example,a ="11"b ="1"Return"100". 题意:将两个以字符串形式保存的二进制数进行相加. 思路:其实不管是以数组.字符串形式,加或者乘(multiply strings),一般的思路都是从后往前计算,用一个中间变量保存相加或者相乘的结果(加法是一个变量int 或string就行,…

C++ class Solution { public: /** * @param a a number * @param b a number * @return the result */ string addBinary(string& a, string& b) { // Write your code here ") { return b; } ") { return a; } string result; ; int ai, bj, sum; char va…