#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)