Divide Two Integers(math)
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.
分析
先都abs,注意 Map: p,q,i,ans =map(abs,(dividend, divisor,0,0))
divident = divisor*(1+2+4+6+8.....)
注意其实<< >>就是 *2和//2。 ans = 1+2+4+6+8.....
Algorithm for integer division of p/q, where p and q are positive:
i, result = 0, 0 # Initialize stuff.
while q << i <= p: i += 1 # Phase 1: Figure out how far left you should go.
for j in reversed(range(i)): # Phase 2: Divide like a 7-year-old.
if q << j <= p:
p -= q << j
result += 1 << j
Time: O(log(answer)) = O(log(dividend // divisor)) = O(log(dividend) - log(divisor))
Space: O(1)
注意这里判断2个异号的方法 :
(dividend<0)!=(divisor<0)
negative = (dividend < 0) ^ (divisor < 0)
if negative:
r = ~r + 1
还有就是int的取值
[-1<<31, 1<<31-1]
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
p,q,i,ans =map(abs,(dividend, divisor,0,0))
while q<<i <= p:
i+=1
for j in reversed(range(i)):
if q<<j <= p:
p -= q<<j
ans+=1<<j
if (dividend<0)!=(divisor<0) or ans < -1<<31:
ans = -ans
return min(ans,(1<<31)-1)
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