Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 这道题属于数值处理的题目,对于整数处理的问题,比较重要的注意点在于符号和处理越界的问题.对于这道题目,因为不能用乘除法和取余运算,我们只能使用位运算和加减法.比较直接的方法是用被除数一直减去除数,直到为0.这种方法的迭代次数是结果的大小,即比如结果为n,算法复杂度是O(n).…

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: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. 由于不能用乘号,除号,和取余.那么一个数除另外一个数,如a/b,实际含义是a中有多少个b,我们可以多次计算a-b,并更新a,最后a-b<0说明循环结束,循环的次数也即结果. 但是上面这种方法超时,如2147483647/1,那么循环次数为2147483647次. 所以考虑二分法来减少循环的次数.而且乘2,相对于左移一位,没有用到乘号. 代码:…

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…

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…