[LeetCode] 67. Add Binary 二进制数相加
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", b = "1011"
Output: "10101"
解法: 从最低位加到最高位,当前位相加结果是%2,进位是/2,记得处理每一次的进位和最后一次的进位,最后反向输出字符。
Java:
public class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1, j = b.length() -1, carry = 0;
while (i >= 0 || j >= 0) {
int sum = carry;
if (j >= 0) sum += b.charAt(j--) - '0';
if (i >= 0) sum += a.charAt(i--) - '0';
sb.append(sum % 2);
carry = sum / 2;
}
if (carry != 0) sb.append(carry);
return sb.reverse().toString();
}
}
Java:
public class AddBinary {
public String addBinary(String a, String b) {
StringBuilder result = new StringBuilder();
int pointerA = a.length()-1;
int pointerB = b.length()-1;
int carry = 0;
while(pointerA>=0 || pointerB>=0){
int sum = carry;
if(pointerA>=0){
sum += (a.charAt(pointerA)-'0');
pointerA--;
}
if(pointerB>=0){
sum += (b.charAt(pointerB)-'0');
pointerB--;
}
result.append(sum%2);
carry = sum/2;
}
if(carry!=0){
result.append('1');
}
return result.reverse().toString();
}
}
Python:
class Solution:
# @param a, a string
# @param b, a string
# @return a string
def addBinary(self, a, b):
result, carry, val = "", 0, 0
for i in xrange(max(len(a), len(b))):
val = carry
if i < len(a):
val += int(a[-(i + 1)])
if i < len(b):
val += int(b[-(i + 1)])
carry, val = val / 2, val % 2
result += str(val)
if carry:
result += str(carry)
return result[::-1]
Python:
class Solution:
def addBinary(self, a, b):
if len(a)==0: return b
if len(b)==0: return a
if a[-1] == '1' and b[-1] == '1':
return self.addBinary(self.addBinary(a[0:-1],b[0:-1]),'1')+'0'
if a[-1] == '0' and b[-1] == '0':
return self.addBinary(a[0:-1],b[0:-1])+'0'
else:
return self.addBinary(a[0:-1],b[0:-1])+'1'
Python:
class Solution(object):
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
res = ''
carry = 0
i, j = len(a) - 1, len(b) - 1
while i >= 0 or j >= 0:
su = carry
if i >= 0:
su += int(a[i])
if j >= 0:
su += int(b[j])
carry = su / 2
res += str(su % 2)
i -= 1
j -= 1 if carry > 0:
res += str(carry) return res[::-1]
C++ 1:
class Solution {
public:
string addBinary(string a, string b) {
string res;
size_t res_len = max(a.length(), b.length()) ; size_t carry = 0;
for (int i = 0; i < res_len; ++i) {
const size_t a_bit_i = i < a.length() ? a[a.length() - 1 - i] - '0' : 0;
const size_t b_bit_i = i < b.length() ? b[b.length() - 1 - i] - '0' : 0;
size_t sum = carry + a_bit_i + b_bit_i;
carry = sum / 2;
sum %= 2;
res.push_back('0' + sum);
}
if (carry) {
res.push_back('0' + carry);
}
reverse(res.begin(), res.end()); return res;
}
};
C++ 2:
class Solution {
public:
string addBinary(string a, string b) {
string res = "";
int m = a.size() - 1, n = b.size() - 1, carry = 0;
while (m >= 0 || n >= 0) {
int p = m >= 0 ? a[m--] - '0' : 0;
int q = n >= 0 ? b[n--] - '0' : 0;
int sum = p + q + carry;
res = to_string(sum % 2) + res;
carry = sum / 2;
}
return carry == 1 ? "1" + res : res;
}
};
Followup:
如果不是二进制相加,而是十六进制相加呢?只要把算法中的除2和余2换成16,并添加相应的十六进制字母就行了。
如果是带小数的?
All LeetCode Questions List 题目汇总
[LeetCode] 67. Add Binary 二进制数相加的更多相关文章
- [LeetCode] Add Binary 二进制数相加
Given two binary strings, return their sum (also a binary string). For example,a = "11"b = ...
- [LintCode] Add Binary 二进制数相加
Given two binary strings, return their sum (also a binary string). Have you met this question in a r ...
- Leetcode 67 Add Binary 大数加法+字符串处理
题意:两个二进制数相加,大数加法的变形 大数加法流程: 1.倒置两个大数,这一步能使所有大数对齐 2.逐位相加,同时进位 3.倒置两个大数的和作为输出 class Solution { public: ...
- LeetCode 67. Add Binary (二进制相加)
Given two binary strings, return their sum (also a binary string). For example,a = "11"b = ...
- LeetCode 67 Add Binary(二进制相加)(*)
翻译 给定两个二进制字符串,返回它们的和(也是二进制字符串). 比如, a = "11" b = "1" 返回 "100". 原文 Give ...
- (String) leetcode 67. Add Binary
Given two binary strings, return their sum (also a binary string). The input strings are both non-em ...
- leetCode 67.Add Binary (二进制加法) 解题思路和方法
Given two binary strings, return their sum (also a binary string). For example, a = "11" b ...
- LeetCode 67. Add Binary
Given two binary strings, return their sum (also a binary string). For example,a = "11"b = ...
- Java [Leetcode 67]Add Binary
题目描述: Given two binary strings, return their sum (also a binary string). For example,a = "11&qu ...
随机推荐
- Celery(异步任务,定时任务,周期任务)
1.什么是Celery Celery是基于Python实现的模块,用于异步.定时.周期任务的. 组成结构: 1.用户任务 app 2.管道broker 用于存储任务 官方推荐 redis/rabbit ...
- [转] C++ explicit关键字详解
本文转自tiankong19999 首先, C++中的explicit关键字只能用于修饰只有一个参数的类构造函数, 它的作用是表明该构造函数是显示的, 而非隐式的, 跟它相对应的另一个关键字是impl ...
- 后台返回的Json为null的字段不显示的方法
如果引入的是谷歌的gson的话,需要引入依赖: <dependency> <groupId>com.fasterxml.jackson.core</groupId> ...
- centos7中,mysql连接报错:1130 - Host ‘118.111.111.111’ is not allowed to connect to this MariaDB server
客户端连接报错 这个问题是因为用户在数据库服务器中的mysql数据库中的user的表中没有权限. 解决步骤 1.连接服务器: mysql -u root -p 2.看当前所有数据库:show data ...
- LeetCode 959. Regions Cut By Slashes
原题链接在这里:https://leetcode.com/problems/regions-cut-by-slashes/ 题目: In a N x N grid composed of 1 x 1 ...
- 微信小程序中登录操作-----与-----引用
login.wxml <view> <!-- <image src="./88.png"></image> --> # 在当前目录下 ...
- H5中实现加载更多的逻辑及代码执行。
H5中加载更多的逻辑总结: 1.首先,需要三个底部的提示,分别是“加载中”.“--我是有底线的--”.“暂时没有记录”,当然,这三句话根据不同的项目,可以自定义.具体代码例子如下: <div c ...
- learning java Paths Path
import java.nio.file.Path; import java.nio.file.Paths; public class PathTest { public static void ma ...
- centos gcc 新版本安装的一种方法
最近出来一个v 语言,打算试用下,但是需要编译,centos 7 gcc 版本太低,一种可选的解决方法 使用Software Collections Software Collections 安装方法 ...
- gj的交换机在升级了ios之后最新数据不刷新,
下午2点开始升级5点结束,之后监控项获取不到最新数据,显示网络接口一直是down的状态,但是登上设备之后显示的是正常up状态, 怀疑是自动发现规则的问题,但是查看之后都是1个小时,应该不会, 这时候诡 ...