实现两整数相除,不容许使用乘法、除法、和取余运算。git
若是结果溢出(int范围为-2147483648 ~ 2147483647),返回MAX_INTgithub
题目原文:ide
Divide two integers without using multiplication, division and mod operator.code
If it is overflow, return MAX_INT..ip
example leetcode
input: 1001, 2 output: 500
利用减法,将被除数减去除数,减去的次数累计即为最后结果get
为了解决效率问题,能够减去除数的倍数,利用位运算,每次除数左移一位(2倍),次数相应加对应的倍数。input
若是左移一位的除数过大,除数还原。it
注意处理除法运算中正负号的问题。io
class Solution(object): def divide(self, dividend, divisor): """ :type dividend: int :type divisor: int :rtype: int """ pos = (dividend < 0) is (divisor < 0) dividend, divisor = abs(dividend), abs(divisor) result = 0 while dividend >= divisor: tmp, i = divisor, 1 while dividend - tmp >= 0: dividend -= tmp result += i i <<= 1 tmp <<= 1 if not pos: result = -result return min(2147483647, max(result, -2147483648))
本题以及其它leetcode题目代码github地址: github地址