Divide Two Integers
视频讲解 http://v.youku.com/v_show/id_XMTY1MTAyODM3Ng==.html
int 范围:
Integer.MIN_VALUE => -2147483648 Integer.MAX_VALUE => 2147483647
overflow:
当被除数为 Integer.MIN_VALUE的时候,取绝对值或者除以-1都会造成溢出overflow.
Math.abs(-2147483648) => -2147483648 Integer.MIN_VALUE/(-1) => -2147483648
(1)把两个数转化成long类型,可以得到正确的绝对值
Long.MAX_VALUE => 9223372036854775807 Long.MIN_VALUE => -9223372036854775808 long a = Math.abs((long)dividend); long b = Math.abs((long)divisor); Math.abs((long)-2147483648) => 2147483648
(2)题目中给出If it is overflow, return MAX_INT.
if(dividend == Integer.MIN_VALUE && divisor == -1)
return Integer.MAX_VALUE;
a=0,b=0, leetcode这道题中这个test case没有处理,所以暂时不需要考虑。
a<b, 被除数小于除数,结果为0。
a>=b, 通过b位移来得到b和a之间的倍数关系。
4<<1 => 4*2,
4<<2 => 4*2*2,
4<<3 => 4*2*2*2,
a=25, b=4, result =0
当a>=b, a>0,b>0
int count =0;记录位移的位数,
如果a> b<<(count+1), count++;
(1) count =0: 25 > 4<<1, 4*2=8 , count++;
count=1: 25 > 4<<2, 4*2*2=16, count++;
count=2: 25 > 4<<3, 4*2*2*2=32不成立,
本轮循环结束,count=2;
result += 1<<count; result =1*2*2=4;
a -= b<<count, a = 25 - 4*2*2 =9;
(2)count=0, 9 > 4<<1, 4*2=8,count++;
count=1, 9 > 4<<2, 4*2*2=16不成立,count=1
本轮结束循环, count=1,
result += 1<<count; result=4+1*2=6;
a -= b<<count, 9-4*2=1, 1<4整个循环结束
由于a和b是被除数和除数的绝对值,如何判断两者符号是否相同?
1^1 => 0
1^0 => 1
true^false => true
true^true => false
false^false => false
((dividend>0)^(divisor>0))?(-result):result
符号相同则为false,取result
不同则为true,取-result的值
public class Solution {
public int divide(int dividend, int divisor) {
if(dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE;
long a = Math.abs((long)dividend);
long b = Math.abs((long)divisor);
if(a<b) return 0;
int result = 0;
while(a>0 && b>0 && a>=b){
int count = 0;
while(a> b<<(count+1)){
count++;
}
result += 1<<count;
a -= b<<count;
}
return ((dividend>0)^(divisor>0))?(-result):result;
}
}
Divide Two Integers的更多相关文章
- [LeetCode] Divide Two Integers 两数相除
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
- Leetcode Divide Two Integers
Divide two integers without using multiplication, division and mod operator. 不用乘.除.求余操作,返回两整数相除的结果,结 ...
- leetcode-【中等题】Divide Two Integers
题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, r ...
- [LintCode] Divide Two Integers 两数相除
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
- 62. Divide Two Integers
Divide Two Integers Divide two integers without using multiplication, division and mod operator. 思路: ...
- Divide Two Integers leetcode
题目:Divide Two Integers Divide two integers without using multiplication, division and mod operator. ...
- Java for LeetCode 029 Divide Two Integers
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
- [LeetCode] Divide Two Integers( bit + 二分法 )
Divide two integers without using multiplication, division and mod operator. 常常出现大的负数,无法用abs()转换成正数的 ...
- LeetCode29 Divide Two Integers
题目: Divide two integers without using multiplication, division and mod operator. If it is overflow, ...
- 【leetcode】Divide Two Integers (middle)☆
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
随机推荐
- 异步-学习笔记3 Task
1. 通过Task启动多线程 2. 解决多线程的几大应用场景 private void btnTask_Click(object sender, EventArgs e) { Console.Writ ...
- SQL Server 2012 启动
1. 启动 SQL Server Management studio 2. 选择登录模式 Server name: "." 代表本地的数据库 Authertication: ...
- Windows10安装MongoDB
环境:Windows10x64,mongodb-win32-x86_64-2008plus-ssl-3.2.9-signed.msi 步骤: 安装msi文件到D:\ 新建配置文件mongo.confi ...
- CentOS7安装Nginx并部署
服务器IP是192.168.36.136 1.直接yum install nginx即可 2.主配置文件是/etc/nginx/下的nginx.conf,另外一个是/etc/nginx/conf.d/ ...
- 多日期选择jQuery插件 MultiDatesPicker for jQuery UI
Multiple-Dates-Picker-for-jQuery-UI是一个多日期选择的jquery控件. GIT源码: https://github.com/dubrox/Multiple-Da ...
- Ruby on rails学习笔记——安装环境
出现问题: C:\Users\len>gem install rails ERROR: While executing gem ... (Gem::RemoteFetcher::FetchErr ...
- C++常用的#include头文件总结
C++常用的#include头文件总结 这篇文章主要介绍了C++常用的#include头文件,对初学者理解C++程序设计大有好处的相关资料 本文详细罗列了C++所包含的头文件的名称及作用说明,比较 ...
- css样式中遇到!important
链接: http://zhidao.baidu.com/link?url=XyYHS2l-bFkzEgMBWfbQYuEV4vmGz8kOhj-jDL_HBqu0KnVIBQJKvw5OOO-bVjv ...
- 基于php使用memcache存储session的详解(转)
web服务器的php session都给memcached ,这样你不管分发器把 ip连接分给哪个web服务器都不会有问题了,配置方法很简单,就在php的配置文件内增加一条语句就可以了,不过前提你需要 ...
- TableView的执行流程 & 位移枚举
// 闲来无聊测试一下 第一轮: 1.numberOfSectionsInTableView :假如section=2,此函数只执行一次,假如section=0,下面函数不执行,默认为1 2.h ...