题目解析：

用加减法实现除法用减法，每次累加被减部分，累加商， 以一个固定的倍数递增

坑： 注意 while循环的跳出便捷，= 的情况要注意。

应用：累加思想，可以用在提速上，效率提高

"""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 = 3Output: 3Example 2:Input: dividend = 7, divisor = -3Output: -2Note:    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."""import timeimport mathclass Solution:    def divide(self, dividend, divisor):        """        :type dividend: int        :type divisor: int        :rtype: int        """        sign=(dividend>0)^(divisor>0)  #如果 ==1，则是 负的 ==0，则是正的        dividend_current=int(math.fabs(dividend))        divisor_current=int(math.fabs(divisor))        quotient=0        quotient_base=1        while dividend_current>0:            print(dividend_current,divisor_current,quotient)            while dividend_current>=divisor_current:                quotient+=quotient_base                dividend_current-=divisor_current                divisor_current*=2   #                quotient_base*=2                # time.sleep(1)            while divisor<=dividend_current
    
     = 2 ** 31:            return 2 ** 31 - 1        elif ans <= -2 ** 31 - 1:            return -2 ** 31        else:            return ansif __name__=='__main__':    st=Solution()    dividend=128    dividend=-2147483648    divisor=1    # out=st.divide(-128,3)    out=st.divide(dividend,divisor)    print(out)