29. Divide Two Integers

1. Description

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, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.

2. Note

Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [$−2^{31}$, $2^{31} − 1$]. For this problem, assume that your function returns $2^{31} − 1$ when the division result overflows.

2. Example

Example 1:
Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.

Example 2:
Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.

Example 3:
Input: dividend = 0, divisor = 1
Output: 0

Example 4:
Input: dividend = 1, divisor = 1
Output: 1

4. Constraints

• $−2^{31}$ <= dividend, divisor <= $2^{31} − 1$
• divisor != 0

5. Solutions

My Accepted Solution

n = m_nums.size()
Time complexity: O(n)
Space complexity: O(1)

// using while loop to decrease divisor is too slow
// so the only option is the bit operation
class Solution {
public:
    int divide(int dividend, int divisor) {
        if(divisor == 1) {
            return dividend;
        } else if(dividend == INT_MIN && divisor == -1) {
            return INT_MAX;
        }

        int positive = (dividend > 0) == (divisor > 0);
        long left = abs(dividend), right = abs(divisor), multiple = 0;
        while(left >= right) {
            int left_move = 0;
            while(left >= (right << left_move)) {
                ++left_move;
            }

            multiple += 1 << (left_move - 1);
            left -= right << (left_move - 1);
        }

        return positive ? multiple : -multiple;
    }
};