图文讲解选择排序算法的原理及在Python中的实现

def sort_choice(numbers, max_to_min=True):
 """
 我这没有按照标准的选择排序，假设列表长度为n，思路如下：
  1、获取最大值x，将x移动到列最后。[n1, n2, n3, ... nn]
  2、将x追加到排序结果[n1, n3, ... nn, n2]
  3、获取排序后n-1个元素[n1, n3, ... nn]，重复第一步，重复n-1次。

 max_to_min是指从大到小排序，默认为true；否则从小到大排序。
 对[8, 4, 1, 0, 9]排序，大致流程如下：
 sorted_numbers = []
 [8, 4, 1, 0, 9], sorted_numbers = [9]
 [4, 1, 0, 8], sorted_numbers = [9, 8]
 [1, 0, 4], sorted_numbers = [9, 8, 4]
 [0, 1], sorted_numbers = [9, 8, 4, 1]
 [0], sorted_numbers = [9, 8, 4, 1, 0]
 """
 if len(numbers) <= 1:
  return numbers
 sorted_list = []
 index = 0
 for i in xrange(len(numbers) - index):
  left_numbers = _get_left_numbers(numbers, max_to_min)
  numbers = left_numbers[:-1]
  sorted_list.append(left_numbers[-1])
  index += 1
 return sorted_list

def _get_left_numbers(numbers, get_max=True):
 '''
 获取最大值或者最小值x，并且将x抽取出来，置于列表最后.
 Ex: get_max=True, [1, 4, 3] ⇒ [1, 3, 4] 
  get_max=False, [1, 4, 3] ⇒ [4, 3 ,1] 
 '''
 max_index = 0
 for i, num in enumerate(numbers):
  if get_max:
   if num > numbers[max_index]:
    max_index = i
  else:
   if num < numbers[max_index]:
    max_index = i
 numbers = numbers[:max_index] + numbers[max_index + 1:] + [numbers[max_index]]
 return numbers

>>> get_left_numbers([0, 4, 0, 31, 9, 19, 89,67], get_max=True)
[0, 4, 0, 31, 9, 19, 67, 89]
>>> get_left_numbers([0, 4, 0, 31, 9, 19, 89,67], get_max=False)
[4, 0, 31, 9, 19, 89, 67, 0]

>>> sort_choice([0, 4, 0, 31, 9, 19, 89,67], max_to_min=False)
[0, 0, 4, 9, 19, 31, 67, 89]
>>> sort_choice([0, 4, 0, 31, 9, 19, 89,67], max_to_min=True)
[89, 67, 31, 19, 9, 4, 0, 0]