Divide Two Integers

Divide Two Integers

问题：

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

If it is overflow, return MAX_INT.

思路：

　　二分查找

　　系统解法：https://blog.csdn.net/whuwangyi/article/details/40995863

我的代码：

public class Solution {
   public int divide(int dividend, int divisor) {
        if(divisor == 0) return Integer.MAX_VALUE;
        if(dividend == 0) return 0;
        boolean flag = true;
        if((dividend < 0 && divisor > 0)||(dividend > 0 && divisor < 0))
        {
            flag = false;
        }
        long top = Math.abs((long)dividend);
        long down = Math.abs((long)divisor);
        long[] rst = helper(top, down);
        long result = (flag ? rst[0] : -rst[0]);
        if(result > Integer.MAX_VALUE || result < Integer.MIN_VALUE)    return Integer.MAX_VALUE;
        return (int)result ;
    }
    public long[] helper(long dividend, long divisor)
    {
        long[] rst = new long[2];
        if(dividend == divisor)
        {
            rst[0] = 1;
            rst[1] = 0;
            return rst;
        }
        if(dividend < divisor)
        {
            rst[0] = 0;
            rst[1] = dividend;
            return rst;
        }
        long mid = dividend >> 1;
        
        rst = helper(mid, divisor);
        long remain = dividend - (mid << 1) + (rst[1] <<1);
        if(remain >= divisor)
        {
            rst[0] = (rst[0]<<1) + 1;
            rst[1] = remain - divisor;
        }
        else
        {
            rst[0] = (rst[0]<<1);
            rst[1] = remain;
        }
        return rst;
    }
}

他人代码：

public int divide(int dividend, int divisor) {
        long a = Math.abs((long)dividend);
        long b = Math.abs((long)divisor);
        
        long ret = 0;
        
        while (a >= b) {
            for (long tmp = b, cnt = 1; a >= tmp; tmp <<= 1, cnt <<= 1) {
                ret += cnt;
                a -= tmp;
            }
        }
        
        ret = (((dividend ^ divisor) >> 31) & 1) == 1 ? -ret: ret;
        if (ret > Integer.MAX_VALUE || ret < Integer.MIN_VALUE) {
            return Integer.MAX_VALUE;
        }
        
        return (int)ret;
    }

学习之处：

• 判断两个数相乘是否为负数， ret = (((dividend ^ divisor) >> 31) & 1) == 1 ? -ret: ret; 这个方法真简洁
• 我的代码太过于臃肿，看看别人的代码是如此的简介，我的代码可取之处只是思路更加清晰一点而已
• 既然不能用乘除法，还可以用移位操作嘛