LeetCode

题目：

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

If it is overflow, return MAX_INT.

思路：

吐槽这种题目，这就像你明明可以买到过年回家的直达票，有票且一键解决，却去买分段票，路途不耽误时间吗？而且买分段票时中间容易出差错(bugs)，最终可能回不了家(运行结果错误)。

package manipulation;

public class DivideTwoIntegers {

    public int divide(int dividend, int divisor) {
        if ((dividend == Integer.MIN_VALUE && divisor == -1) || divisor == 0) return Integer.MAX_VALUE;
        if (divisor == 1) return dividend;
        if (divisor == -1) return -dividend;
        
        long a = Math.abs((long)dividend);
        long b = Math.abs((long)divisor);
        
        long base = b;
        int i = 0;
        long[] nums = new long[32];
        nums[0] = b;
        while (nums[i] <= a && nums[i] > 0) {
            nums[i+1] = base << (i+1);
            ++i;
        }
        --i;
        
        int cnt = 0;
        while (a > 0 && i >= 0) {
            while (a - nums[i] >= 0) {
                a -= nums[i];
                cnt += (1 << i);
            }
            --i;
        }
        
        return ((dividend ^ divisor) >> 31) == 0 ? cnt : -cnt;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        DivideTwoIntegers d = new DivideTwoIntegers();
        System.out.println(d.divide(-2147483648, 2));
    }

}