Divide Two Integers

This question is very hard, just remember it:
Explaination:

思路清晰，就是二倍法。
直接用除数去一个一个加，直到被除数被超过的话，会超时。
解决办法每次将被除数增加1倍，同时将count也增加一倍，如果超过了被除数，那么用被除数减去当前和再继续本操作。因为一个很大的数经过对数处理后都能在可接受的大小内。比如Integer.MAX_VALUE 的log后也不过31.
另外这道题很恶心的一点是在边界的处理，因为我们知道最大允许的负数的绝对值比最大允许的正数要大1. 所以以后遇到这类可能越界的问题，一律用比它大的集合来装。。

public class Solution {
    /**
     * @param dividend the dividend
     * @param divisor the divisor
     * @return the result
     */
    public int divide(int dividend, int divisor) {
        // Write your code here
        if (divisor == 0) {
            return dividend > 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
        }
        if (dividend == 0) {
            return 0;
        }
        //This is the situation where the number is out of bound;
        //Integer.MIN_VALUE = -2147483648
        //Integer.MAX_VALUE = 2147483647
        if (dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }
      
        long a = Math.abs((long) dividend);
        long b = Math.abs((long) divisor);
        
        
        int cnt = 0;
        while (a >= b) { //must have equal because if it is equal, we can still deduct
            int shift = 0;
            while (a >= (b << shift)) { // equal b * 2^shift
                shift++;
            }
            a -= b << (shift - 1);
            cnt += 1 << (shift - 1); 
            //We need the third if because if it doesn't exist, the 1 will be shift to the first bit, which will change the cnt to Integer.MIN_VALUE
        }
        
        int rst = (((dividend ^ divisor) >> 31) & 1) == 1? -cnt : cnt;
        
        return (int)rst;
    }
}