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
- springMVC的注解详解
springmvc常用注解标签详解 1.@Controller 在SpringMVC 中,控制器Controller 负责处理由DispatcherServlet 分发的请求,它把用户请求的数据经过业 ...
- CIQRCodeGenerator Core Image Filter Reference
https://developer.apple.com/library/prerelease/content/documentation/GraphicsImaging/Reference/CoreI ...
- tomcat十大安全优化措施
1.telnet管理端口保护 使用telnet连接进来可以输入SHUTDOWN可以直接关闭tomcat,极不安全,必须关闭.可以修改默认的管理端口8005改为其他端口,修改SHUTDOWN指令为其他字 ...
- 药企信息sop
中国药品生产企业 http://db.yaozh.com/shengchanqiye 全球药品生产企业 http://db.yaozh.com/quanqiuqiye
- SpringMVC中的设计模式
1.<跟我学SpringMVC> P10 2.<跟我学SpringMVC> P32
- R树空间索引及其变种
1.R树及其变种:百度百科 2.R树详介:http://blog.csdn.net/jazywoo123/article/details/7792745 3.R树及变种小结 R树:叶子节点或中间节点都 ...
- Linux版MonoDevelop无法连接调试器的解决方案(Could not connet to the debugger)
安装了Linux版本的MonoDevelop之后,在运行程序的时候会提示Could not connnet to the debugger.的错误. 原因是新版本的Gnome Terminal不再接受 ...
- Order Independent Transparency
http://on-demand.gputechconf.com/gtc/2014/presentations/S4385-order-independent-transparency-opengl. ...
- JabRef 文献管理软件
JabRef 文献管理软件简明教程 大多只有使用LaTeX撰写科技论文的研究人员才能完全领略到JabRef的妙不可言,但随着对Word写作平台上BibTeX4Word插件的开发和便利应用,使用Word ...
- maven可选依赖(Optional Dependencies)和依赖排除(Dependency Exclusions)
我们知道,maven的依赖关系是有传递性的.如:A-->B,B-->C.但有时候,项目A可能不是必需依赖C,因此需要在项目A中排除对A的依赖.在maven的依赖管理中,有两种方式可以对依赖 ...