LeetCode_29. Divide Two Integers 不用乘、除、模运算来实现两个数的除法

题目：

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.

 

代码：

class Solution(object):
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        
        sign = 1               #标记符号
        if dividend > 0 and divisor < 0  or dividend < 0 and divisor > 0 :
            sign = -1
        
        dividend, divisor = abs(dividend), abs(divisor)
        
        result = 0     #最终结果
        
        #保留两个数，mi_div为实际除数，每除一次，mi_div翻一倍；add_time为商，跟mi_div保持同步
        mi_div = divisor       
        add_time = 1
        
        while dividend >= divisor :
            if dividend >= mi_div :       #如果被除数大于实际除数
                dividend -= mi_div        #减法代替除法
                result += add_time        #结果加上除数的倍数
                mi_div = mi_div << 1      #除数继续翻倍
                add_time = add_time << 1  #商翻倍
            else :
                mi_div = mi_div >> 1      #否则，当前被除数小于实际除数了，但仍大于原始除数，实际除数缩小
                add_time = add_time >> 1
        
        return max(-2**31, min(sign*result, 2**31-1))
        
        
#for example, if we want to calc (17/2)

# ret = 0;

# 17-2 ,ret+=1; left=15

# 15-4 ,ret+=2; left=11

# 11-8 ,ret+=4; left=3

# 3-2 ,ret+=1; left=1

# ret=8;