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…

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…