好吧,我有这个比较合并排序和选择排序的python代码,但它需要永远。当从n = 0到90,000(列表的大小)完成时,对列表进行排序仅需要大约3秒。通过这个逻辑,它将需要大约10 * 3 * 9秒(运行次数*持续时间*递增的运行[我们从10,000开始然后做20,000,然后是30,000等])。但是,它需要更长的时间。
import time
import random
# Selection Sort Code #
def maxIndex(J):
return J.index(max(J))
def swap(LCopy, i, j):
temp = LCopy[i]
LCopy[i] = LCopy[j]
LCopy[j] = temp
# Implementation of selection sort
def selectionSort(L):
for i in range(len(L)-1, 1, -1):
j = maxIndex(L[0:i+1])
swap(L, i, j)
# Merge Sort Code #
# Assumes that L[first:mid+1] is sorted and also
# that L[mid: last+1] is sorted. Returns L with L[first: last+1] sorted
def merge(L, first, mid, last):
i = first # index into the first half
j = mid + 1 # index into the second half
tempList = []
# This loops goes on as long as BOTH i and j stay within their
# respective sorted blocks
while (i <= mid) and (j <= last):
if L[i] <= L[j]:
tempList.append(L[i])
#print L[i], "from the first block"
i += 1
else:
tempList.append(L[j])
#print L[j], "from the second block"
j += 1
# If i goes beyond the first block, there may be some elements
# in the second block that need to be copied into tempList.
# Similarly, if j goes beyond the second block, there may be some
# elements in the first block that need to be copied into tempList
if i == mid + 1:
tempList.extend(L[j:last+1])
#print L[j:last+1], "some elements in second block are left over"
elif j == last+1:
tempList.extend(L[i:mid+1])
#print L[i:mid+1], "some elements from first block are left over"
L[first:last+1] = tempList
#print tempList
# The merge sort function; sorts the sublist L[first:last+1]
def generalMergeSort(L, first, last):
# Base case: if first == last then it is already sorted
# Recursive case: L[first:last+1] has size 2 or more
if first < last:
# divide step
mid = (first + last)/2
# conquer step
generalMergeSort(L, first, mid)
generalMergeSort(L, mid+1, last)
# combine step
merge(L, first, mid, last)
# Wrapper function
def mergeSort(L):
generalMergeSort(L, 0, len(L)-1)
m = 10
n = 100000
n_increments = 9
baseList = [ random.randint(0,100) for r in range(n) ]
i = 0
while i < n_increments:
j = 0
sel_time = 0
mer_time = 0
while j < m:
# Do a Selection Sort #
x = time.clock()
selectionSort( baseList)
y = time.clock()
sel_time += ( y - x )
random.shuffle( baseList )
# Do a Merge Sort #
x = time.clock()
mergeSort( baseList )
y = time.clock()
mer_time += ( y - x )
random.shuffle( baseList )
j += 1
print "average select sort time for a list of", n, "size:", sel_time / m
print "average merge sort time for a list of", n, "size:", mer_time / m
j = 0
i += 1
n += 10000
答案 0 :(得分:4)
因为您正在使用O(n ^ 2)排序算法。这意味着如果你加倍n,算法运行时间要长4倍。请注意,您的起点是100,000而不是10,000