题目
You are given an array nums of n positive integers.
You can perform two types of operations on any element of the array any number of times:
- If the element is even, divide it by 2.
- For example, if the array is [1,2,3,4], then you can do this operation on the last element, and the array will be [1,2,3,2].
- If the element is odd, multiply it by 2.
- For example, if the array is [1,2,3,4], then you can do this operation on the first element, and the array will be [2,2,3,4]. The deviation of the array is the maximum difference between any two elements in the array.
Return the minimum deviation the array can have after performing some number of operations.
解题思路
将奇数*2，和所有偶数一起放入优先队列，每次pop出最大的，和最小的做差更新结果，最大的是奇数时结束循环
代码
class Solution:
def minimumDeviation(self, nums: List[int]) -> int:
q = []
for n in nums:
q.append(-2 * n if n & 1 else -n)
res, maxv = float('inf'), max(q)
heapq.heapify(q)
while True:
t = heapq.heappop(q)
res = min(res, maxv - t)
if t & 1:
break
t >>= 1
maxv = max(maxv, t)
heapq.heappush(q, t)
return res
