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;
}
}
};