Python程序:查找子列表中大小至少为k的最大平均值
假设我们有一个名为nums的数字列表和另一个值k,我们需要找到长度至少为k的任何子列表的最大平均值。
因此,如果输入类似于nums = [2, 10, -50, 4, 6, 6] k = 3,则输出将为5.33333333,因为子列表[4, 6, 6]具有最大的平均值。
为了解决这个问题,我们将遵循以下步骤:
left := nums的最小值,right := nums的最大值
s := nums中从索引0到k-1的所有数字的和
largest_avg := s / k
当left <= right时,执行以下操作:
mid := (left + right) / 2的整数部分
sum1 := s,avg := s / k,sum2 := 0,cnt := 0
对于范围k到nums大小的i,执行以下操作:
sum1 := sum1 + nums[i]
sum2 := sum2 + nums[i - k]
cnt := cnt + 1
avg := avg和(sum1 / (cnt + k))的最大值
如果sum2 / cnt <= mid,则:
sum1 := sum1 - sum2
cnt := 0,sum2 := 0
avg := avg和(sum1 / (cnt + k))的最大值
largest_avg := largest_avg和avg的最大值
如果avg > mid,则:
left := mid + 1
否则:
right := mid - 1
返回largest_avg
让我们看一下下面的实现,以便更好地理解:
示例
class Solution: def solve(self, nums, k): left, right = min(nums), max(nums) s = sum(nums[:k]) largest_avg = s / k while left <= right: mid = (left + right) // 2 sum1 = s avg = s / k sum2 = 0 cnt = 0 for i in range(k, len(nums)): sum1 += nums[i] sum2 += nums[i − k] cnt += 1 avg = max(avg, sum1 / (cnt + k)) if sum2 / cnt <= mid: sum1 −= sum2 cnt = 0 sum2 = 0 avg = max(avg, sum1 / (cnt + k)) largest_avg = max(largest_avg, avg) if avg > mid: left = mid + 1 else: right = mid − 1 return largest_avg ob = Solution() nums = [2, 10, −50, 4, 6, 6] k = 3 print(ob.solve(nums, k))
输入
[2, 10, −50, 4, 6, 6], k = 3
Learn Python in-depth with real-world projects through our Python certification course. Enroll and become a certified expert to boost your career.
输出
5.333333333333333