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给定两个整数,被除数 dividend 和除数 divisor。将两数相除,要求不使用乘法、除法和 mod 运算符。
返回被除数 dividend 除以除数 divisor 得到的商。
示例 1:
输入: dividend = 10, divisor = 3 输出: 3 示例 2:
输入: dividend = 7, divisor = -3 输出: -2 说明:
被除数和除数均为 32 位有符号整数。 除数不为 0。 假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−231, 231 −
1]。本题中,如果除法结果溢出,则返回 231 − 1。
这道题考察的位运算的除法。首先我们会想到用加法或者减法来代替除法,代码如下:
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
divide_num = 0
if dividend > 0 and divisor > 0 or dividend <= 0 and divisor <= 0:
while dividend >= divisor:
dividend -= divisor
divide_num += 1
return divide_num
else:
if dividend < 0:
dividend = -dividend
elif divisor < 0:
divisor = -divisor
while dividend >= divisor:
dividend -= divisor
divide_num += 1
return -divide_num
if __name__ == "__main__":
s = Solution()
dividend = 1
divisor = 1
print(s.divide(dividend,divisor))
显然这样的代码会超时
于是我们自然而然会想到用一种更简洁的除法来代替
于是想到用位运算除法
不懂可以参考:https://www.jianshu.com/p/7bba031b11e7
代码如下:
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
result = 0
temp = 0
if divisor >= 0 and dividend >= 0 or divisor < 0 and dividend < 0:
sign = 1
else:
sign = -1
if dividend == 0 or divisor == 0:
return 0
dividend = abs(dividend)
divisor = abs(divisor)
for i in range(32, -1, -1):
if temp + (divisor << i) <= dividend:
temp += divisor << i
result |= 1 << i
if sign < 0:
result = -result
elif sign > 0:
result = result
if result < -2**31 or result > (2**31) -1:
return 2**31 -1
else:
return result
if __name__ == "__main__":
s = Solution()
dividend = 2**31 -1
divisor = 1
print(s.divide(dividend,divisor))