#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：

 1 class Solution:
 2     # @return an integer
 3     def dividePositive(self, dividend, divisor):
 4         if dividend < divisor:
 5             return 0
 6
 7         sum = divisor
 8         count = 1
 9         while sum + sum < dividend:
10             sum += sum
11             count += count
12
13         count += self.dividePositive(dividend - sum, divisor)
14
15         return count
16
17     def divide(self, dividend, divisor):
18         if dividend >= 0:
19             if divisor > 0:
20                 return self.dividePositive(dividend, divisor)
21             else:
22                 return 0 - self.dividePositive(dividend, 0 - divisor)
23         else:
24             if divisor > 0:
25                 return 0 - self.dividePositive(0 - dividend, divisor)
26             else:
27                 return self.dividePositive(0 - dividend, 0 - divisor)

#Leet Code# Divide Two Integers