#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 =…

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: dividend…

题目描述: Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解题思路: 把除数表示为:dividend = 2^i * divisor + 2^(i-1) * divisor + ... + 2^0 * divisor.这样一来,我们所求的商就是各系数之和了,而每个系数都可以通过移位操作获得. 详细解说请参考:http:/…

  Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: divide…

转载:https://blog.csdn.net/Lynn_Baby/article/details/80624180 Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer divi…

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: dividend…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解法: 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. 采用位运算中的移位运算,左移一位相当于乘2,右移一位相当于除以2.假设求 a / b,将b左移n位后大于a,则结果 res += 1 << (n - 1),将a更新 (a -= b <<…

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: dividend…