算法39--Divide Two Integers

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.

实现除法，不能使用乘除以及取余运算

最初思路是先确定最终符号，然后转换为两个数的绝对值进行运算，从被除数中不断减去除数，直到被除数小于除数停止，返回最终结果：

class Solution:
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        if dividend==0:
            return 0
        flag = True
        if dividend*divisor<0:
            flag = False
        dd = abs(dividend)
        ds = abs(divisor)
        if dd<ds:
            return 0
        num = 1
        tmp = dd-ds
        while tmp>ds:
            num += 1
            tmp = tmp - ds
        if flag:
            return num
        else:
            return -num

后来发现超时了：

实现除法，不能使用乘除以及取余运算，如果继续优化可能会使用位运算

换一个角度思考，我们一次取除法的整数次倍数，直到该值刚好不大于被除数，此时倍数即为最终答案，

假设被除数dd 除数ds 商为num 余数为x

有dd=ds*num+x (0<=x<ds) 现在求num的值，考虑num的二进制，假设有n位Mn-1Mn-2....M1M0，即

num=Mn-1*pow(2,n-1)+.....+M1*pow(2,1)+M0*pow(2,0)

先求出n-1的值利用ds不断左移直到刚好不大于dd，此时移动的位数即为n-1,dd=dd-ds*pow(2,n-1)继续执行上述过程求出n-2。。。。

最终结果有：num=Mn-1*pow(2,n-1)+.....+M1*pow(2,1)+M0*pow(2,0)

class Solution:
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        maxv = pow(2,31)-1
        minv = -pow(2,31)
        if dividend==0:
            return 0
        flag = True
        if dividend*divisor<0:
            flag = False
        dd = abs(dividend)
        ds = abs(divisor)        
        num = 0
        tmp = ds
        while dd>=ds:
            m = 1
            tmp = ds
            while dd>(tmp<<1):
                tmp = tmp << 1
                m = m << 1
            num += m
            dd -= tmp        
        if flag:
            return num if num<=maxv else maxv
        else:
            return -num if -num>=minv else minv

利用移位运算来模拟某个数

算法39--Divide Two Integers

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