#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 =…

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…

1. Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the…

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…