#include<bits/stdc++.h> using namespace std; int divide(int dividend, int divisor) { long long n = dividend, m = divisor; // determine sign of the quotient ^ m < ? - : ; // remove sign of operands n = abs(n), m = abs(m); // q stores the quotient…

描述 不能使用乘法.除法和取模(mod)等运算,除开两个数得到结果,如果内存溢出则返回Integer类型的最大值.解释一下就是:输入两个数,第一个数是被除数dividend,第二个是除数divisor,要求是在不得使用乘法.除法和取模(mod)等运算的前提下,求出两个数的相除结果. 思路 有一个最简单直观的方法,设置一个i=1,比较dividend和divisor大小,如果满足dividend>divisor,就令divisor=divisor+divisor,i++,继续判断dividend>…

class Solution(object): def divide(self, dividend, divisor): """ :type dividend: int :type divisor: int :rtype: int """ ispositive = True and divisor < : ispositive = False and divisor > : ispositive = False dividend =…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 这道题让我们求两数相除,而且规定我们不能用乘法,除法和取余操作,那么我们还可以用另一神器位操作Bit Operation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更…

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,结果也是整数. 假设除数是2,相除的商就是被除数二进制表示向右移动一位. 假设被除数是a,除数是b,因为不知道a除以b的商,所以只能从b,2b,4b,8b.......这种序列一个个尝试 从a扣除那些尝试的值. 如果a大于序列的数,那么a扣除该值,并且最终结果是商加上对应的二进制位为1的数,然后尝试序…

题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT 链接 https://leetcode.com/problems/divide-two-integers/ 答案 1.int的最大值MAX_INT为power(2,31)-1 = 2147483647 2.int的最小值MIN_INT为-power(2,31) = -21…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return 2147483647 Have you met this question in a real interview?     Example Given dividend = 100 and divisor = 9, return 11. LeetCode上的原题,请参见我之前的博客Divid…

Divide Two Integers Divide two integers without using multiplication, division and mod operator. 思路: 类同 趣味算法之数学问题:题4. 两点需要注意: 1. 除数或被除数为最大负数时,转化为正数会溢出.2. divisor + divisor 可能会溢出. class Solution { public: int divide(int dividend, int divisor) { if(div…

题目:Divide Two Integers Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 看讨论区大神的思路: In this problem, we are asked to divide two integers. However, we are not allowed to use division, multi…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解题思路: 既然不呢个用乘除和取模运算,只好采用移位运算,可以通过设置一个length代表divisor向左的做大移位数,直到大于dividend,然后对length做一个循环递减,dividend如果大于divisor即进行减法运算,同时result加上对应的值,注意边界条…