Divide two integers without using multiplication, division and mod operator. 思路:1.先将被除数和除数转化为long的非负数,注意一定要为long.由于Integer.MIN_VALUE的绝对值超出了Integer的范围. 2.常理:不论什么正整数num都能够表示为num=2^a+2^b+2^c+...+2^n.故能够採用2^a+2^b+2^c+...+2^n来表示商,即dividend=divisor*(2^a+2^…

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…

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. 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:/…

  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…

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…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解法: 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. 采用位运算中的移位运算,左移一位相当于乘2,右移一位相当于除以2.假设求 a / b,将b左移n位后大于a,则结果 res += 1 << (n - 1),将a更新 (a -= b <<…

Divide two integers without using multiplication, division and mod operator. class Solution { public: int divide(int dividend, int divisor) { ? -(long long)dividend : dividend; ? -(long long)divisor : divisor; && divisor >= ) || (dividend <=…