相當於複雜度優化到logN.
如 100 3
普通方法是 3 + 3 + 3 + 3 … + 3累加到100，數加了多少次
優化方法 3 6 12 24…
1 2 4 8… 看能累加多少次，超過的方法減去求得剩下的值，繼續進行累加。
注意邊界條件

package 牛客刷題.LeetCode.模擬.divide_two_integers;

/**
 * Created by Administrator on 2018/5/26 0026.
 */
public class Solution {
    public static void main(String[] args) {
        //System.out.println(divide(Integer.MIN_VALUE, -1));
 

        //System.out.println(divide(8, 1));
        //System.out.println(divide(24,3));
        System.out.println(divide(-1010369383 ,-2147483648));
    }

    public static int divide(int dividend, int divisor) {
        if (dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }
        long
 
 a = Math.abs((long)dividend);
        long b = Math.abs((long)divisor);
        int num = 0;
        long sum;
        while (b <= a) {
            sum = b;
            int count = 1;
            while (sum + sum <= a) {
                count += count;
                sum += sum;
            }
            a = a - sum;
            num = num + count;
        }
        if
 
 ((dividend < 0 && divisor > 0) || dividend > 0 && divisor < 0) {
            num = -num;
        }
        return num;
    }
}