LeetCode - Divide Two Integers

Divide Two Integers

2013.12.7 04:52

Divide two integers without using multiplication, division and mod operator.

Solution:

　　Another bit-manipulation problem, integer division. Normally if we calculate x / y (x, y > 0), we'll subtract y from x until y < x. This operation can be accelerated if we subtract x << (..., 2, 1, 0) from y, until x < y, using only log2(x) time.

　　Time complexity is O(log(dividend) - log(divisor)), space complexity is O(1).

Accepted code:

// 2RE, 2WA, 1AC
class Solution {
public:
    int divide(int dividend, int divisor) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        // 2RE here, -2147483648 is a tricky case
        
        long long int div1, div2;
        int s1, s2;
        long long int base;
        // 1WA here, $r overflows, must be long long int
        long long int res, r;
        
        s1 = sign(dividend);
        s2 = sign(divisor);
        
        if(s1 * s2 == 0){
            return 0;
        }
        
        div1 = dividend > 0 ? dividend : -(long long int)dividend;
        div2 = divisor > 0 ? divisor : -(long long int)divisor;
        
        base = div2;
        r = 1;
        while(base <= div1){
            base <<= 1;
            r <<= 1;
        }
        
        res = 0;
        while(r > 0){
            if(div1 >= base){
                div1 -= base;
                res += r;
            }
            base >>= 1;
            r >>= 1;
        }
        
        if(s1 < 0){
            res = -res;
        }
        if(s2 < 0){
            res = -res;
        }
        
        // 1WA here, forgot to multiply sign variables
        return res;
    }
private:
    int sign(int x){
        if(x > 0){
            return 1;
        }else if(x < 0){
            return -1;
        }else{
            return 0;
        }
    }
};