python 利用quick sort思路实现median函数

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# import numpy as np   def median(arr):     #return np.median(arr)     arr.sort()     return arr[len(arr)>>1]     def patition(arr, low, high):     pivot = arr[low]     i = low+1     while i <= high:         if arr[i] > pivot:             arr[i], arr[high] = arr[high], arr[i]             high -= 1         else:             i += 1     arr[high], arr[low] = arr[low], arr[high]     return high     def median_helper(arr, low, high, found_index):     pivot_index = patition(arr, low, high)     if pivot_index == found_index:         return arr[found_index]     elif pivot_index > found_index:         return median_helper(arr, low, pivot_index - 1, found_index)     else:         return median_helper(arr, pivot_index + 1, high, found_index)     def median2(arr):     assert arr     mid = len(arr)>>1     return median_helper(arr, 0, len(arr) - 1, mid)     from random import randint for j in range(2, 2000):     arr = [randint(0, 2000) for i in range(1, j)]     a = median(list(arr))     b = median2(list(arr))     if a != b:         print(a)         print(b)         print("debug:")         a = median(list(arr))         b = median2(list(arr))

 时间复杂度：O（2n）

因为 n+n/2+n/4+.... = 2n

下面这样写也很直观，比我写的跑起来还快些（诡异）：

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import random def quick_select(A, k):     #pivot value is random     pivot = random.choice(A)       A1 = [] #values < pivot     A2 = [] #values > pivot       for i in A:         if i < pivot:             A1.append(i)         elif i > pivot:             A2.append(i)         else:             pass  # ignore Pivot value!       #case 1: median is in A1     if k <= len(A1):         return quick_select(A1, k)     #case 2: median is in A2     elif k > len(A) - len(A2):         return quick_select(A2, k - (len(A) - len(A2)))     #case 3: median found     else:         return pivot

 

C=n+n2+n4+n8+…=2n=O(n)