Python笔试题目：求最大的K个数子，解法二，适合小型数据集的情况

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题目：

Givena array of 10,000 random intergers, select the biggest 100 numbers.

1)The order of the result numbers does not matter;

2)Take care about the algorithm performance and big O complexity.

我的解答：

#coding=utf-8## generate random numbersfrom random import randint# low and high limit of the numbers of the random numberlow = -10000000high = 10000000# total_number of the numberstotal_number = 10000# the number of beggest number we needmax_number = 100# use  () for [] will be more efficient ?numbers = [randint(low,high) for elem in xrange(total_number)]#print numbers"""   when the dataset is not large, we still consider the method of "sort and then select"   take quick sort for example, its average complexity is O(N*logN),   then take the K beggest numbers with complexity O(K). So    time complexity is O(N*logN) + O(K) = O(N*logN) in total.   If we only find the biggest K numbers and let the N-K number alone,   the complexity is O(N*K) using part_sorting.   comparing O(N*logN) and O(N*k), we can find that quick sort algorithm is more efficient that the latter one    when k > logN and vice versa. """from math import logdef  quick_sort(numbers):        return [] if numbers == [] else quick_sort([y for y in numbers[1:] if y < numbers[0]]) +  \           [numbers[0]] + quick_sort([y for y in numbers[1:] if y >= numbers[0]]) def selection_sort_part(numbers):    size = len(numbers)    # the range is [0...max_number-1] ,rather than [0,len(numbers)]. so the selection performs only K times,    for i in range(max_number):          k = i         for j in range(i + 1, size):            #               if numbers[j] > numbers[k]:                  k = j                            if k is not i:              numbers[i], numbers[k] = numbers[k], numbers[i]                return numbersdef main():    if max_number >= log(total_number,2):        print 'result from quick_sort algorithm:',quick_sort(numbers)[-max_number:]    else:        print 'result from selection_sort_part algorithm:',selection_sort_part(numbers)[:max_number]if __name__ == '__main__':    main()

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