我决定最近学习python!我想使用以下代码编写一个简单的合并排序:
def mergeSort(lst):
l = len(lst)
if l <= 0:
print("empty")
return None
elif l == 1:
return lst
half = int(l / 2)
m = lst[half]
print(half, m)
left = []
right = []
for n in lst:
if n < m:
left.append(n)
else:
right.append(n)
left = mergeSort(left)
right = mergeSort(right)
return merge(left, right)
不幸的是,当代码必须处理诸如[1 1 1]之类的列表时,此代码会生成无限循环。你能建议一些方法来解决这个错误的行为吗?
答案 0 :(得分:4)
你签出了http://www.geekviewpoint.com/吗?这可能是学习如何用Python编写算法的最佳方法。看看这个。作为奖励,它是一个非常干净的网站,我最近看到的唯一广告是关于axdlab的一个名为“Puzz!”的android智能拼图应用程序。该网站本身有各种算法和良好的解释。
以下是他们的合并排序:
#=======================================================================
# Author: Isai Damier
# Title: Mergesort
# Project: geekviewpoint
# Package: algorithm.sorting
#
# Statement:
# Given a disordered list of integers (or any other items),
# rearrange the integers in natural order.
#
# Sample Input: [8,5,3,1,9,6,0,7,4,2,5]
#
# Sample Output: [0,1,2,3,4,5,5,6,7,8,9]
#
# Time Complexity of Solution:
# Best = Average = Worst = O(nlog(n)).
#
# Approach:
# Merge sort is a divide and conquer algorithm. In the divide and
# conquer paradigm, a problem is broken into pieces where each piece
# still retains all the properties of the larger problem -- except
# its size. To solve the original problem, each piece is solved
# individually; then the pieces are merged back together.
#
# For illustration, imagine needing to sort an array of 200 elements
# using selection sort. Since selection sort takes O(n^2), it would
# take about 40,000 time units to sort the array. Now imagine
# splitting the array into ten equal pieces and sorting each piece
# individually still using selection sort. Now it would take 400
# time units to sort each piece; for a grand total of 10400 = 4000.
# Once each piece is sorted, merging them back together would take
# about 200 time units; for a grand total of 200+4000 = 4,200.
# Clearly 4,200 is an impressive improvement over 40,000. Now
# imagine greater. Imagine splitting the original array into
# groups of two and then sorting them. In the end, it would take about
# 1,000 time units to sort the array. That's how merge sort works.
#
# NOTE to the Python experts:
# While it might seem more "Pythonic" to take such approach as
#
# mid = len(aList) / 2
# left = mergesort(aList[:mid])
# right = mergesort(aList[mid:])
#
# That approach take too much memory for creating sublists.
#=======================================================================
def mergesort( aList ):
_mergesort( aList, 0, len( aList ) - 1 )
def _mergesort( aList, first, last ):
# break problem into smaller structurally identical pieces
mid = ( first + last ) / 2
if first < last:
_mergesort( aList, first, mid )
_mergesort( aList, mid + 1, last )
# merge solved pieces to get solution to original problem
a, f, l = 0, first, mid + 1
tmp = [None] * ( last - first + 1 )
while f <= mid and l <= last:
if aList[f] < aList[l] :
tmp[a] = aList[f]
f += 1
else:
tmp[a] = aList[l]
l += 1
a += 1
if f <= mid :
tmp[a:] = aList[f:mid + 1]
if l <= last:
tmp[a:] = aList[l:last + 1]
a = 0
while first <= last:
aList[first] = tmp[a]
first += 1
a += 1
这是测试平台:
import unittest
from algorithms import sorting
class Test( unittest.TestCase ):
def testMergesort( self ):
A = [8, 5, 3, 1, 9, 6, 0, 7, 4, 2, 5]
sorting.mergesort( A )
for i in range( 1, len( A ) ):
if A[i - 1] > A[i]:
self.fail( "mergesort method fails." )
答案 1 :(得分:2)
我相信你只是应该在中间点将列表分成两半 - 而不是将每个项目分成两半。
所以不要这样:
left = []
right = []
for n in lst:
if n < m:
left.append(n)
else:
right.append(n)
这样做:
left = lst[:half]
right = lst[half:]
答案 2 :(得分:1)
您实施的算法是(有缺陷的)快速排序,而不会删除所谓的“pivot”元素,在您的情况m
中。
您执行的合并操作不需要像合并排序那样进行任何合并,因为如果您要正确处理数据透视表,对mergeSort(left)
的调用将返回已排序的left
。
在合并排序中,您没有透视元素m
,相反,您只需将列表分为两部分,如James所述。
根据经验,递归调用应始终在较小的数据集上运行。