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…

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…

class Solution { public int divide(int dividend, int divisor) { if (dividend == Integer.MIN_VALUE && divisor == -1) { return Integer.MAX_VALUE; } long a = Math.abs((long)dividend); long b = Math.abs((long)divisor); int num = 0; long sum; while (b…

给定两个整数,被除数 dividend 和除数 divisor.将两数相除,要求不使用乘法.除法和 mod 运算符. 返回被除数 dividend 除以除数 divisor 得到的商. 示例 1: 输入: dividend = 10, divisor = 3 输出: 3 示例 2: 输入: dividend = 7, divisor = -3 输出: -2 说明: 被除数和除数均为 32 位有符号整数. 除数不为 0. 假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−231,  2…

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…

如题,不用乘除法和mod实现两数相除. 这里引用一位clever boy 的解法. class Solution { public: int divide(int dividend, int divisor) { ; ) ; ) return INT_MAX; res = exp(log(llabs(dividend)) - log(llabs(divisor))); ) ^ (divisor < )) res = -res; if(res > INT_MAX) res = INT_MAX;…

本文始发于个人公众号:TechFlow,原创不易,求个关注 链接 Divide Two Integers 难度 Medium 描述 给定两个整数,被除数和除数,要求在不使用除号的情况下计算出两数的商 Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividi…

难受啊 考虑越界 考虑dividend为-2^31,用负数移位运算 class Solution { public: int divide(int dividend, int divisor) { if(dividend == divisor) return 1; if(divisor == INT_MIN) return 0; if(dividend==INT_MIN&&divisor==-1) return INT_MAX; if(dividend==INT_MIN&&…