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
- BZOJ3678: wangxz与OJ
splay缩点. #include<bits/stdc++.h> #define L(t) (t)->c[0] #define R(t) (t)->c[1] #define F ...
- python学习笔记-(六)深copy&浅copy
在python中,对象赋值实际上是对象的引用.当创建一个对象,然后把它赋给另一个变量的时候,python并没有拷贝这个对象,而只是拷贝了这个对象的引用. 1. 赋值 赋值其实只是传递对象引用,引用对象 ...
- Socket通信的理解
1.Socket(套接字) 是支持TCP/IP通信的基本操作单元.包含通信的五种必须信息:通信使用的协议,本机IP和端口,远程IP和端口. 2. 1.TCP连接 手机能够使用联网功能是因为手机底层实现 ...
- logo上传
- JSP 属性范围
参考文献:http://www.cnblogs.com/xdp-gacl/p/3781056.html 一.属性范围 所谓的属性范围就是一个属性设置之后,可以经过多少个其他页面后仍然可以访问的保存范围 ...
- Linux下Redis服务器安装配置
说明:操作系统:CentOS1.安装编译工具yum install wget make gcc gcc-c++ zlib-devel openssl openssl-devel pcre-devel ...
- Intent启动一个新的页面
一,Intent(目的) 的分类 显式 Intent 构造函数重载之一: Intent intent = new Intent(FirstActivity.this,SecondActivity.cl ...
- centos忘记开机密码
系统:centos6.6,忘记开机密码,进入单用户模式进行重置,以下为操作过程. 1. reset(重启)Linux系统,在出现如下图的界面时,请点Enter键,确保一定要快,只存在几秒.. 2.点击 ...
- vi/vim使用小结
1.三种模式: •Command mode 命令模式,用于输入命令,简单更改. •Insert mode 插入模式,用于插入文本. •Last line mode 末行模式,用于输入命令.视化操作.查 ...
- Redis-cluster集群【第四篇】:redis-cluster集群配置
Redis分片: 为什么要分片:随着Redis存储的数据越来越庞大,会导致Redis的性能越来越差! 目前分片的方法: 1.客户端分片 在应用层面分片,程序里指定什么数据存放在那个Redis 优势: ...