描述:不使用 * / % 完成除法操作.O(n)复杂度会超时,需要O(lg(n))复杂度. 代码: class Solution: # @return an integer def dividePositive(self, dividend, divisor): if dividend < divisor: return 0 sum = divisor count = 1 while sum + sum < dividend: sum += sum count += count count +…

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…

  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…

Divide Two Integers Divide two integers without using multiplication, division and mod operator. SOLUTION 11. 基本思想是不断地减掉除数,直到为0为止.但是这样会太慢. 2. 我们可以使用2分法来加速这个过程.不断对除数*2,直到它比被除数还大为止.加倍的同时,也记录下cnt,将被除数减掉加倍后的值,并且结果+cnt. 因为是2倍地加大,所以速度会很快,指数级的速度. 3. 另外要注意的是…

Divide Two Integers 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…

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…

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…

Divide two integers without using multiplication, division and mod operator. 常常出现大的负数,无法用abs()转换成正数的情况 class Solution{ private: vector<long long> f; public: int bsearch(vector<long long> &a,int left,int right,long long key){ if(left > r…

题目: Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. (Medium) 分析: 题目要求不使用乘除和模运算实现两个整数除法. 第一个思路就是每次把count加等被除数自身判定,只到count<=除数,并且count + 被除数 > 除数时即为结果. 但是考虑到可能有 MAX_INT / 1这种情况,肯定华丽超时. 然后…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 思路: 尼玛,各种通不过,开始用纯减法,超时了. 然后用递归,溢出了. 再然后终于开窍了,用循环,把被除数每次加倍去找答案,结果一遇到 -2147483648 就各种不行, 主要是这个数一求绝对值就溢出了. 再然后,受不了了,看答案. 发现,大家都用long long来解决溢…

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:/…

leetcode面试准备:Divide Two Integers 1 题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 接口: public int divide(int dividend, int divisor) 2 思路 题意 不用乘.除.mod 做一个除法运算. 解 直接用除数去一个一个加,直到被除数被超过的话…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 题目大意:不用乘除取模运算计算两个数的除. 解题思路:只能用位运算了,当被除数大于除数,除数左移1位.2位……直到得到最大的然后用被除数减去它,将因数加到res上,这里有点二分的意思,依次循环往复,这里我把两个数都设为负数,因为0x80000000是最小的负数,它没有对应的最…

Question Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. Solution dividend = divisor * quotient + remainder 而我们知道对于任何一个数可以表示为Σi * 2x  其中i为0或1.所以我们可以用加法实现乘法. a = a + a 等同于 a = a * 2 因此我们可…

otal Accepted: 54356 Total Submissions: 357733 Difficulty: Medium 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…

# -*- 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. =…

Divide two integers without using multiplication, division and mod operator. 分析:题目意思很容易理解,就是不用乘除法和模运算求来做除法,很容易想到的一个方法是一直做减法,然后计数,超时.在网上找到一种解法,利用位运算,意思是任何一个整数可以表示成以2的幂为底的一组基的线性组合,即num=a_0*2^0+a_1*2^1+a_2*2^2+...+a_n*2^n.基于以上这个公式以及左移一位相当于乘以2,我们先让除数左移直到…

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: 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…

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:             Inp…

leetcode-algorithms-29 Divide Two Integers 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 trunc…

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…

题目: Divide two integers without using multiplication, division and mod operator. 题解: 这道题我自己没想出来...乘除取模都不让用..那只有加减了...我参考的http://blog.csdn.net/perfect8886/article/details/23040143 代码如下:                       }                                          …

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…