29 Divide Two Integers

29. Divide Two Integers

1. Question

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

If it is overflow, return MAX_INT.

2. Implementation

(1) 倍增法

思路: 我们知道每个数都可以用二进制来表示，为了快速得到两个数相除的结果，我们可以利用倍增法的方式得到两个数相除的商的二进制形式。举个例子，87 / 4 = 21 (忽略余数部分), 而 21的二进制形式是10101，所以 87 = 4 * (1 * 2 ^ 4 + 0 * 2 ^ 3 + 1 * 2 ^ 2 + 0 * 2 ^ 1 + 1 * 2 ^ 0)

class Solution {
    public int divide(int dividend, int divisor) {
        long a = Math.abs((long)dividend);
        long b = Math.abs((long)divisor);
        long res = 0;

        while (a >= b) {
            long temp = b;
            int i = 0;

            while (a >= temp) {
                a -= temp;
                temp <<= 1;
                res += 1 << i;
                ++i;
            }
        }

        if (dividend > 0 && divisor < 0 || dividend < 0 && divisor > 0) {
            res = -res;
        }
        return res > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int)res;
    }
}

3. Time & Space Complexity

倍增法:时间复杂度O(log(divisor)), 空间复杂度O(1)