Divide Two Integers

public 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.
        if(dividend == 0)
            return 0;
        if(divisor == 1)
            return dividend;
        boolean negative = false;
        if(dividend < 0 && divisor > 0)
            negative = true;
        if(dividend > 0 && divisor < 0)
            negative = true;
        long[] mydivisor = new long[32];
        long ldivisor = (long)Math.abs((long)divisor);
        long ldividend = (long)Math.abs((long)dividend);
        int i = 0;
        while(i < 32)
        {
            if(ldividend >= ldivisor)
            {
                mydivisor[i] = ldivisor;
                ldivisor <<= 1;
            }
            else
            {
                i--;
                break;
            }
            i++;
        }
        
        int result = 0;
        while(i>=0)
        {
            if(ldividend - mydivisor[i] >= 0)
                {
                    ldividend -= mydivisor[i];
                    result += (1<<i);
                }
            else
                {
                    i--;
                }
        }
        if(negative)
            return ~result + 1;
        else
            return result;
    }
}

两个数的除法，但是不允许用乘、除、取余符号。

[解题思路]
如果可以用乘的话，二分搜索倒是不错的解法。
否则，只能寄希望于位符操作了。

 基本思想就是把除数向左移位(×2)然后与被除数比较，直到发现仅次于被除数的那个值，减去该值后继续。也可以用递归做，这里图省事，就是一个循环了事。

ref:http://fisherlei.blogspot.com/2012/12/leetcode-divide-two-integers.html