python3实现数据结构与算法30天-排序-选择排序(5)

算法思路:
每次选出最小的，放入第一个位置，
再一趟排序记录列表无序区最小的数，放第二个位置

算法关键点：有序区和无序区，无序区最小的数
时间复杂度：O(n^2)

1.简单版-代码:

import

def select_sort_simple(lst):
    """create new list, callback min & remove"""
    lst_new = []
    for i in range(len(lst)):
        min_loc = min(lst)
        lst_new.append(min_loc)
        lst.remove(min_loc)
        print(lst_new)
    return lst_new


lst = [random.randint(0, 100) for x in range(10)]
print(lst)
print(select_sort_simple(lst))

结果:

[33, 67, 92, 15, 39, 23, 10, 53, 95, 25]
[10]
[10, 15]
[10, 15, 23]
[10, 15, 23, 25]
[10, 15, 23, 25, 33]
[10, 15, 23, 25, 33, 39]
[10, 15, 23, 25, 33, 39, 53]
[10, 15, 23, 25, 33, 39, 53, 67]
[10, 15, 23, 25, 33, 39, 53, 67, 92]
[10, 15, 23, 25, 33, 39, 53, 67, 92, 95]
[10, 15, 23, 25, 33, 39, 53, 67, 92, 95]

上面方法可以说是选择排序的简单实现，但存在几个问题，min(),remove()其实也是O(n)级别，而且开辟新的列表，消耗内存，如果序列很大，内存资源消耗大，不建议用。
下面的相当于升级版，改善之处在于，不用开辟新的序列，另外做到能抠则抠，尽量减少不必要的循环，选择排序推荐使用。

2.升级版-代码:

import random

def select_srot(lst):
    for i in range(len(lst)-1): # i是第几趟
        min_loc = i
        for j in range(i+1, len(lst)): # j是无序区, 从i+1开始
            if lst[min_loc] > lst[j]:
                min_loc = j
        lst[i], lst[min_loc] = lst[min_loc], lst[i]
        print(lst)

lst = [random.randint(0, 100) for x in range(10)]
print(lst)
print(select_srot(lst))

结果:

[3, 22, 92, 84, 97, 52, 27, 39, 1, 63]
[1, 22, 92, 84, 97, 52, 27, 39, 3, 63]
[1, 3, 92, 84, 97, 52, 27, 39, 22, 63]
[1, 3, 22, 84, 97, 52, 27, 39, 92, 63]
[1, 3, 22, 27, 97, 52, 84, 39, 92, 63]
[1, 3, 22, 27, 39, 52, 84, 97, 92, 63]
[1, 3, 22, 27, 39, 52, 84, 97, 92, 63]
[1, 3, 22, 27, 39, 52, 63, 97, 92, 84]
[1, 3, 22, 27, 39, 52, 63, 84, 92, 97]
[1, 3, 22, 27, 39, 52, 63, 84, 92, 97]