29. 两数相除

 

难度中等966收藏分享切换为英文接收动态反馈

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。

整数除法的结果应当截去（truncate）其小数部分，例如：truncate(8.345) = 8 以及 truncate(-2.7335) = -2

 

示例 1:

输入: dividend = 10, divisor = 3
输出: 3
解释: 10/3 = truncate(3.33333..) = truncate(3) = 3

示例 2:

输入: dividend = 7, divisor = -3
输出: -2
解释: 7/-3 = truncate(-2.33333..) = -2

class Solution {
public:
    long mul(long a, long k) {
        long ans = 0;
        while (k > 0) {
            if ((k&1) == 1) ans += a; //if ((k%2) == 1) ans += a;
            k = k >> 1; //k = k/2;
            a += a;
        }
    return ans;
}
    int divide(int dividend_1, int divisor_1) {
        bool neg = false;
        long long dividend = dividend_1;
        long long divisor = divisor_1;
        if ((dividend^divisor) < 0) {
            neg = true;
        }
        if (divisor < 0) {
            divisor = -divisor;
           
        }
        if (dividend < 0) {
            dividend = -dividend;
           
        }
        long long low = 0;
        long long high = dividend;
        while(low<=high) {
            long long mid = low + (high-low)/2;
            //cout << mid << endl;
            if (mul(mid,divisor) > dividend) {
                high = mid -1;
            } else if (mul(mid,divisor) < dividend) {
                low = mid +1;
            } else {
                long res  = neg?-mid:mid;
                 if(res > INT_MAX) res = INT_MAX;
                if(res <= INT_MIN) res = INT_MIN;
                return res;
            }
        }
        int res = low-1;
        if(neg) res = -res;
        //cout << INT_MIN <<endl;
        if(res > INT_MAX) return INT_MAX;
        if(res <= INT_MIN) return INT_MAX;
        return res;
    }
};