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.

二货当然直接想到的是逐个相减

显而易见的超时了QAQ

想到可能操作和快速幂有点像

但是快速幂是底数一直扩大啊

然后就没深想 看了题解

MD 果不其然

底数先设置成除数，尝试能否整除，能的话翻倍，发现某一步底数太大了，好，还是一半一半的降下去

class Solution(object):
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        if(divisor==0):
            return 0
        flag=1
        if((dividend>0 and divisor<0) or (dividend<0 and divisor>0)):
            flag=-1
        dividend,divisor=abs(dividend),abs(divisor)
        tot,p,q=0,1,divisor
        while(dividend>=divisor):
            if(dividend>=q):
                tot=tot+p
                dividend=dividend-q
                p=(p<<1)
                q=(q<<1)
            else:
                p=(p>>1)
                q=(q>>1)
        return min(max(-2147483648,tot*flag),2147483647);