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…

Question 29. Divide Two Integers Solution 题目大意:给定两个数字,求出它们的商,要求不能使用乘法.除法以及求余操作. 思路:说下用移位实现的方法 7/3=2,7是被除数,3是除数 除数左移,假设移动了n次后,移到最接近被除数,这时被除数=被除数-除数,商的一部分为2^n 如果被除数>除数,则继续循环 除数左移,又移动了m次后,移到最接近被除数,这时被除数=被除数-除数,商的一部分为2^m 最后商为2^n+2^m+... Java实现: 法1:如果可以用除…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 29: Divide Two Integershttps://oj.leetcode.com/problems/divide-two-integers/ Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT. =…

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: dividend…

一天一道LeetCode系列 (一)题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. (二)解题 这题看起来很简单,一开始想到的方法就是从0开始一次累加除数,一直到比被除数大为止,好无悬念,这样做的结果就是超时了. 用移位来实现除法效率就比较高了.具体思路可以参考二进制除法.下面举个例子来说明. 例如:10/2 即10…

  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. 思路I:做减法,直到被除数<除数.但结果 Time Limit Exceeded class Solution { public: int divide(int dividend, int divisor) { ) //分母为0 return INT_MAX; long in…