我在python中的排序算法有时会在运行时冻结,有人可以看看吗?

时间:2016-11-13 13:02:33

标签: python algorithm python-3.x sorting

基本上我一直在尝试的是,我从未排序的列表中挑选出最小和最大的,然后将它们附加到一个新列表,然后从旧的未排序列表中弹出最小和最大的并执行该过程一遍又一遍,直到我最终得到一个排序列表。请看一下我的代码。

import random
import time

stack = [] #sorted list
numbersarray = [] #unsorted list

usersize = int(input("How many digits do you want in your array? ")) #numberofinputs
limit = 0
counter = 0


while limit <= usersize:
    numbersarray.append(random.randint(0,20)) #randomly input numbers into array
    limit = limit + 1

print(numbersarray) #prints the current unsorted array
start_time = time.time() #starts clock

subtractor = 0 #used later in code for changing index
while len(numbersarray) != 0:
    i = 0
    largest = numbersarray[i]
    size = len(numbersarray) -1
    smallest = numbersarray[i]

while (i < len(numbersarray)): 
    if numbersarray[i] >= largest:
        largest = numbersarray[i]
        index = i
    elif numbersarray[i] <= smallest:
        smallest = numbersarray[i]
        indextwo = i
    i = i+1

if (len(numbersarray) == 1): #this checks if there's only 1 number left.
    entry = int(stacksize/2 + 1)
    stack.insert(entry,numbersarray[0])
    break
else:
    if indextwo > index:
        numbersarray.pop(indextwo)
        numbersarray.pop(index)
    elif index > indextwo:
        numbersarray.pop(index)
        numbersarray.pop(indextwo)

stacksize = len(stack)
if stacksize == 0:
    stack.append(smallest)
    stack.append(largest)
elif stacksize != 0:
    stack.insert(stacksize-subtractor,largest) #the subtractor is dynamically changing the index of insertion.
    stack.insert(0+subtractor,smallest)
subtractor = subtractor + 1

print(stack)
print("--- %s seconds ---" % (time.time() - start_time))

1 个答案:

答案 0 :(得分:1)

你有一个双重嵌套的while循环,你在其中扫描maxmin的整个数组,然后从任意位置的列表中继续.pop个元素/ em>的

考虑到pop对于不在列表末尾的项目具有O(N)复杂度;你的方法非常低效,并且会为usersize的大值传递/冻结。这就是为什么,我猜,当usersize很大时,标题中的“有时”会发生。

简而言之,您需要找到一个更好的算法来解决您的问题。