二刷29. Divide Two Integers

    NOTES ON SOME BIT HACKS: 
    1. x << n is equivalent to x * Math.pow(2, n). (<< is Leftshift Operator)
    2. 1 << x is equivalent to Math.pow(2, x). (<< is Leftshift Operator)
    3. x and y have opposite signs if (x ^ y) < 0. (^ is Bitwise XOR)

注意几点，一开始被除数、除数、商全部转化成long. 并且被除数除数都取转化成long之后的绝对值带入计算。需要处理的edge cases只有除数为0，以及被除数为Integer.MIN_VALUE除数为 -1的情况，此时商为Integer.MAX_VALUE. sign都可一开始就求出来。

class Solution {
    public int divide(int dividend, int divisor) {
        if (divisor == 0){
            return Integer.MAX_VALUE;
        }            
        if (dividend == Integer.MIN_VALUE && divisor == -1){
            return Integer.MAX_VALUE;
        }
        int sign = 1;
        if (dividend > 0 && divisor < 0 || (dividend < 0 && divisor > 0)){
            sign = -1;
        }
        long lDividend = Math.abs((long) dividend);
        long lDivisor = Math.abs((long)divisor);
        int powerOfTwo = 0;
        long quotient = 0;
        while (lDividend >= lDivisor){
            powerOfTwo = 0;
            while (lDivisor << powerOfTwo <= lDividend){
                powerOfTwo++;
            }
            powerOfTwo = powerOfTwo - 1;
            quotient += (1 << powerOfTwo);
            lDividend -= (lDivisor << powerOfTwo);
        }
        return (int) quotient*sign;
    }
}