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 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 Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 给定一个整型数组,每个数都出现了三次(只有一个数只出现了一次),找到…
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. 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…
Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Hide Tags Bit Manipulation 数组中的数均出现3…
class Solution(object): def divide(self, dividend, divisor): """ :type dividend: int :type divisor: int :rtype: int """ flag=-1 if ( dividend > 0 and divisor >0 ) or ( dividend < 0 and divisor < 0 ): flag=1 divide…
题意:先将0, 1, 2, ..., n放入数组,然后去掉其中一个值,找到那个值. 这题与singe number 是一个类型,变形的地方就是首先需要将0, 1, 2, ..., n再次放入这个数组,这样就和singe number 一样. class Solution { public: int missingNumber(std::vector<int>& nums) { ; ; i < nums.size(); ++i){ ans ^= nums[i]; } ; i <…
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…
昨天在leetcode做题的时候做到了371,原题是这样的: 371. Sum of Two Integers Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Example: Given a = 1 and b = 2, return 3. 因为之前完全没有在实际练习中使用过位运算,所以刚看到这道题目的时候我的第一反应是 1.用乘除代替加减,但是一想,…
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…
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…
136. Single Number 因为A XOR A = 0,且XOR运算是可交换的,于是,对于实例{2,1,4,5,2,4,1}就会有这样的结果: (2^1^4^5^2^4^1) => ((2^2)^(1^1)^(4^4)^(5)) => (0^0^0^5) => 5异或:异为1 137. Single Number II(黑人问号脸) 对于除出现一次之外的所有的整数,其二进制表示中每一位1出现的次数是3的整数倍,将所有这些1清零,剩下的就是最终的数.用ones记录到当前计算的变量…
题目:两整数相除 难度:Medium 题目内容: 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. 翻…
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…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解题思路: 模拟除法运算,但是不能试用 * / % 操作. 先思考边界条件,再开始做题: 0.除数为0无意义,需要和出题人沟通此边界是否存在(leetcode中并未考察除数为0的情况,默认除数不为0): 1.传入被除数/除数两个int做除法,什么情况返回值overflow:被…
1.题目大意 Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 给定一个数组的整数,数组中的每个元素都出现了两次.例外地,有一个元素只出现…
题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 分析 题目要求不用 * / %三种运算符的条件下,求得两个int类型整数的商. 方法一: 很明显的,我们可以用求和累计的方法,求得商,但是该方法测试会出现TLE:参考博客提出解决办法:每次将被除数增加1倍,同时将count也增加一倍,如果超过了被除数,那么用被除数减去当…
