# Description

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

If it is overflow, return MAX_INT.

The original problem is here.
The original code is here.

# My Solution

I solve this problem in C++, as below:

``````#include<iostream>
#include<math.h>
using namespace std;

class Solution {
public:
int divide(int dividend, int divisor) {
int res  = 0, move = 0;
int MIN_VALUE = -2147483648;
int MAX_VALUE = 2147483647;
if(divisor == 0)
return MAX_VALUE;

if(dividend == MIN_VALUE)
{
if(divisor == -1)
return MAX_VALUE;

dividend += abs(divisor);
res ++;
}
if(divisor == MIN_VALUE)
{
return res;
}
int flag = sign(dividend,divisor);
dividend = abs(dividend);
divisor  = abs(divisor);
int divid = dividend >> 1;
int divis = divisor;
cout<<"dividend="<<divid<<endl;
cout<<"divis="<<divis<<endl;
while(divid >= divis)
{
move +=1;
divis = divis << 1;

}
while(move >= 0)
{
if(dividend >= divis)
{
res += 1<<move;
dividend =  dividend - divis;
}
divis = divis >> 1;
move --;
}

if(flag == 1)
return res;
else
return -res;
}
int sign(int num1, int num2)
{
if((num1 > 0 && num2>0) || (num1 < 0 && num2<0))
return 1;
else
return -1;
}
};

int main()
{
Solution s;
//int dividend = 2147483647;
int dividend = 9;
int divisor  = 1;
int res = s.divide(dividend,divisor);
cout<<res<<endl;
//int a = dividend << 1;
//cout<<"a="<<a<<endl;
//cout<<"dividend="<<dividend<<endl;
}
``````

# Note

At first, we should judge whether the dividend and divisor are Max Value or Min Value. If divisor is zero, directly return the Max value.
Because we can’t use multiplication, division and mod operator, so we use bit and minus operator.