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# Divide Two Integers(math)

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.
```

分析

**先都abs,注意 Map**： p,q,i,ans =map(abs,(dividend, divisor,0,0))

divident = divisor\*(1+2+4+6+8.....)

注意其实<< >>就是 \*2和//2。 ans = 1+2+4+6+8.....

```
Algorithm for integer division of p/q, where p and q are positive:

i, result = 0, 0 # Initialize stuff.

while q << i <= p: i += 1 # Phase 1: Figure out how far left you should go.

for j in reversed(range(i)): # Phase 2: Divide like a 7-year-old.
    if q << j <= p:
        p -= q << j
        result += 1 << j
```

Time: O(log(answer)) = O(log(dividend // divisor)) = O(log(dividend) - log(divisor))

Space: O(1)

**注意这里判断2个异号的方法 ：**

`(dividend<0)!=(divisor<0)`

```
negative = (dividend < 0) ^ (divisor < 0)
if negative:
            r = ~r + 1
```

**还有就是int的取值**

```
[-1<<31, 1<<31-1]
```

```
class Solution:
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        p,q,i,ans =map(abs,(dividend, divisor,0,0))
        while q<<i <= p:
            i+=1

        for j in reversed(range(i)):
            if q<<j <= p:
                p -= q<<j
                ans+=1<<j

        if (dividend<0)!=(divisor<0) or ans < -1<<31:
            ans = -ans

        return min(ans,(1<<31)-1)
```