#Leet Code# Divide Two Integers

描述：不使用 * / % 完成除法操作。O(n)复杂度会超时，需要O(lg(n))复杂度。

代码：

 class Solution:
     # @return an integer
     def dividePositive(self, dividend, divisor):
         if dividend < divisor:
             return 0

         sum = divisor
         count = 1
         while sum + sum < dividend:
             sum += sum
             count += count

         count += self.dividePositive(dividend - sum, divisor)

         return count

     def divide(self, dividend, divisor):
         if dividend >= 0:
             if divisor > 0:
                 return self.dividePositive(dividend, divisor)
             else:
                 return 0 - self.dividePositive(dividend, 0 - divisor)
         else:
             if divisor > 0:
                 return 0 - self.dividePositive(0 - dividend, divisor)
             else:
                 return self.dividePositive(0 - dividend, 0 - divisor)