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));    }}