[Math]Divide Two Integers
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
1.除法转换为加法,加法转换为乘法,y个x相加的结果就是x*y.
假设结果为dividend的一半,用此值与divisor相乘,如果相乘结果大于dividend则,high=dividend,然后继续二分
/**
两个整数相除也可能会溢出,最小负数除以-1就溢出了
*/
class Solution {
public:
long long int multiply(long long x,long long int y)
{
if(y==){
return x;
}
long long int res = multiply(x,y>>);
res = (&y) ? (res<<) + x: (res<<);
return res;
} int divide(int dividend, int divisor) {
long long int l_dividend = fabs(dividend);
long long int l_divisor = fabs(divisor); if(l_dividend < l_divisor){
return ;
}
if((dividend==INT_MIN && divisor==-) || divisor==){
return INT_MAX;
} int sign = ((dividend>>)^(divisor>>)) ? -:; long long int low = ,high =l_dividend; while(low<=high){
long long int mid = low+((high-low)>>);
long long int res = multiply(l_divisor,mid);
if(res == l_dividend){
return mid*sign;
}else if(res<l_dividend){
low = mid+;
}else{
high = mid-;
}
} return (low-)*sign;
}
};
/**
两个整数相除也可能会溢出,最小负数除以-1就溢出了
*/
class Solution {
public:
int divide(int dividend, int divisor) {
long long int l_dividend = fabs(dividend);
long long int l_divisor = fabs(divisor); if(l_dividend < l_divisor){
return ;
}
if((dividend==INT_MIN && divisor==-) || divisor==){
return INT_MAX;
} int sign = ((dividend>>)^(divisor>>)) ? -:;
int res = ;
while(l_dividend >= l_divisor){
long long int tmp = l_divisor;
int occur_times = ;
while(tmp <= l_dividend){
tmp = tmp<<;
occur_times++;
}
res += (<<(occur_times-));
l_dividend -= (tmp>>);
} return res*sign;
}
};
[Math]Divide Two Integers的更多相关文章
- Java for LeetCode 029 Divide Two Integers
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
- Java [leetcode 29]Divide Two Integers
题目描述: Divide two integers without using multiplication, division and mod operator. If it is overflow ...
- leetcode面试准备:Divide Two Integers
leetcode面试准备:Divide Two Integers 1 题目 Divide two integers without using multiplication, division and ...
- Divide Two Integers 解答
Question Divide two integers without using multiplication, division and mod operator. If it is overf ...
- 29. Divide Two Integers (JAVA)
Given two integers dividend and divisor, divide two integers without using multiplication, division ...
- [LeetCode] 29. Divide Two Integers(不使用乘除取模,求两数相除) ☆☆☆
转载:https://blog.csdn.net/Lynn_Baby/article/details/80624180 Given two integers dividend and divisor, ...
- Divide Two Integers leetcode java
题目: Divide two integers without using multiplication, division and mod operator. 题解: 这道题我自己没想出来...乘除 ...
- LeetCode: Divide Two Integers 解题报告
Divide Two Integers Divide two integers without using multiplication, division and mod operator. SOL ...
- [LeetCode] 29. Divide Two Integers ☆☆
Divide two integers without using multiplication, division and mod operator. If it is overflow, retu ...
随机推荐
- SQL Server事务的存储过程
在酒店管理系统开发中,我们会创建房间表和房间类型表(房型表)这两个表,如下图所示: 房型表:RoomType 房间表:Room 首先这两个表的关系:Room是从表,RoomType是主表,两表有主外键 ...
- alarm函数可以定时
貌似是可以的,不过感觉好像这样用不是很好,最好还是用回timer_settimer一些列函数吧,不过既然开了头,就看下alarm怎么用吧. 1. 所需头文件 #include<unistd.h ...
- mysql 数据库连接池
hibernate配置C3P0详解 分类: hibernate 2013-08-14 16:16 1213人阅读 评论 ...
- 使用Promise规定来处理ajax请求的结果
ajax()返回结果是成功的,调用done()中的回调函数: 失败则调用fail()中的回调函数; always()的回调函数不管成功是否都会调用: 可以是使用then()函数代替done()和fai ...
- JQuery中的省市联动
<!DOCTYPE html> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <m ...
- Android SqLite升级
android开发中,如果大家使用到了sqlite就会牵涉到它的升级问题,因为升级后的表结构可能完全不一样,会有字段的添加或者删除等.. sqlite升级思路: 1:将表A重新命名:例 ...
- mysqldump 的方式来搭建master-->slave 的复制架构
1.master 上要满足的最小条件: 1.server_id 已经设置成了一个非0值 2.log_bin 配置好binlog 2.slave 上要满足的最小条件 1.server_id 已经设置成了 ...
- abiword Namespace List
abiword Namespace List Here is a list of all namespaces with brief descriptions: abicollab 这个命名空间以及 ...
- git diff 使用
1. 本地工作目录与远程仓库对比(所有改动过的文件) git diff HEAD (HEAD指向最新一次的提交,即最新版本) 2. 之对比给定的文件 git diff -- filename //是 ...
- DeflateStream类
DeflateStream是另外一种压缩与解压缩流,使用方法与GZipStream类似,而且压缩之后的带下也差不多. 一.属性 BaseStream 获取对基础流的引用. CanRead 获取一个值 ...