Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3

Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Note:

• Both dividend and divisor will be 32-bit signed integers.
• The divisor will never be 0.
• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹,  2³¹ − 1]. For the purpose of this problem, assume that your function returns 2³¹ − 1 when the division result overflows.

题目要求不使用乘除和取余的运算得到除法的结果，我们第一个想到的肯定是减法，除法的本质就是减法嘛，但是减法的时间复杂度很不友好，如果被除数是Integer.MAX_VALUE，除数是1，那就很麻烦了

在这个基础上我们想到了更高级一点的操作方法，位移运算，<<1表示把字节左移一位相当于*2，我们不断扩大除数divisor，直到它再扩大于被除数，这个时候我们需要缩小被除数，就用这个被除数减去当前的除数然后再重复之前的过程，就得到了我们想要的结果。这里需要注意的是测试样例存在integer.MAX_VALUE和-1的特殊情况，所以我们要使用long来帮助存储判断

class Solution {

    public int divide(int dividend, int divisor) {

        long m=Math.abs((long)dividend);

        long n=Math.abs((long)divisor);

        long res=0,tag=Integer.MAX_VALUE;

        if(m<n) return 0;

        while(m>=n){

            long temp=n,count=1;

            while(m>(temp<<1)){

                temp<<=1;

                count<<=1;

            }

            m-=temp;

            res+=count;

        }

        if((dividend<0)^(divisor<0)) res=-res;

        return (int)(res>tag ? tag:res);

    }

}