5.11. 归并排序
.. Copyright (C) Brad Miller, David Ranum This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/4.0/.
The Merge Sort ~~~~~~~~~~~~~~
We now turn our attention to using a divide and conquer strategy as a
way to improve the performance of sorting algorithms. The first
algorithm we will study is the merge sort. Merge sort is a recursive
algorithm that continually splits a list in half. If the list is empty
or has one item, it is sorted by definition (the base case). If the list
has more than one item, we split the list and recursively invoke a merge
sort on both halves. Once the two halves are sorted, the fundamental
operation, called a merge, is performed. Merging is the process of
taking two smaller sorted lists and combining them together into a
single sorted new list. :ref:Figure 10 <fig_mergesortA> shows our familiar example
list as it is being split by merge_sort. :ref:Figure 11 <fig_mergesortB> shows
the simple lists, now sorted, as they are merged back together.
.. _fig_mergesortA:
.. figure:: Figures/mergesortA.png :align: center
Figure 10: Splitting the List in a Merge Sort
.. _fig_mergesortB:
.. figure:: Figures/mergesortB.png :align: center
Figure 11: Lists as They Are Merged Together
The merge_sort function shown in :ref:ActiveCode 1 <lst_mergeSort> begins by asking the
base case question. If the length of the list is less than or equal to
one, then we already have a sorted list and no more processing is
necessary. If, on the other hand, the length is greater than one, then
we use the Python slice operation to extract the left and right
halves. It is important to note that the list may not have an even
number of items. That does not matter, as the lengths will differ by at
most one.
.. _lst_merge:
.. activecode:: lst_mergeSort :caption: Merge Sort
def merge_sort(a_list):
print("Splitting", a_list)
if len(a_list) > 1:
mid = len(a_list) // 2
left_half = a_list[:mid]
right_half = a_list[mid:]
merge_sort(left_half)
merge_sort(right_half)
i, j, k = 0, 0, 0
while i < len(left_half) and j < len(right_half):
if left_half[i] <= right_half[j]:
a_list[k] = left_half[i]
i = i + 1
else:
a_list[k] = right_half[j]
j = j + 1
k = k + 1
while i < len(left_half):
a_list[k] = left_half[i]
i = i + 1
k = k + 1
while j < len(right_half):
a_list[k] = right_half[j]
j = j + 1
k = k + 1
print("Merging", a_list)
a_list = [54, 26, 93, 17, 77, 31, 44, 55, 20]
merge_sort(a_list)
print(a_list)
Once the merge_sort function is invoked on the left half and the
right half (lines 8–9), it is assumed they are sorted. The rest of the
function (lines 11–29) is responsible for merging the two smaller sorted
lists into a larger sorted list. Notice that the merge operation places
the items back into the original list (a_list) one at a time by
repeatedly taking the smallest item from the sorted lists. Note that the
condition in line 13 (left_half[i] <= right_half[j]) ensures that the algorithm is
stable. A stable algorithm maintains the order of duplicate items in
a list and is preferred in most cases.
The merge_sort function has been augmented with a print statement
(line 2) to show the contents of the list being sorted at the start of
each invocation. There is also a print statement (line 30) to show
the merging process. The transcript shows the result of executing the
function on our example list. Note that the list with 44, 55, and 20
will not divide evenly. The first split gives [44] and the second gives
[55, 20]. It is easy to see how the splitting process eventually yields a
list that can be immediately merged with other sorted lists.
.. animation:: merge_anim :modelfile: sortmodels.js :viewerfile: sortviewers.js :model: MergeSortModel :viewer: BarViewer
.. For more detail, CodeLens 6 allows you to step through the algorithm. .. .. .. .. codelens:: mergetrace .. :caption: Tracing the Merge Sort .. .. def merge_sort(a_list): .. print("Splitting ",a_list) .. if len(a_list)>1: .. mid = len(a_list)//2 .. left_half = a_list[:mid] .. right_half = a_list[mid:] .. .. merge_sort(left_half) .. merge_sort(right_half) .. .. i, j, k = 0, 0, 0 .. while i<len(left_half) and j<len(right_half): .. if left_half[i]<=right_half[j]: .. a_list[k]=left_half[i] .. i=i+1 .. else: .. a_list[k]=right_half[j] .. j=j+1 .. k=k+1 .. .. while i<len(left_half): .. a_list[k]=left_half[i] .. i=i+1 .. k=k+1 .. .. while j<len(right_half): .. a_list[k]=right_half[j] .. j=j+1 .. k=k+1 .. print("Merging ",a_list) .. .. a_list = [54,26,93,17,77,31,44,55,20] .. merge_sort(a_list) .. print(a_list)
In order to analyze the merge_sort function, we need to consider the
two distinct processes that make up its implementation. First, the list
is split into halves. We already computed (in a binary search) that we
can divide a list in half :math:\log{n} times where :math:n is the
length of the list. The second process is the merge. Each item in the
list will eventually be processed and placed on the sorted list. So the
merge operation which results in a list of size :math:n requires :math:n
operations. The result of this analysis is that :math:\log{n} splits,
each of which costs :math:n for a total of :math:n\log{n}
operations. A merge sort is an :math:O(n\log{n}) algorithm.
Recall that the slicing operator is :math:O(k) where :math:k is the size
of the slice. In order to guarantee that merge_sort will be
:math:O(n\log n) we will need to remove the slice operator. Again,
this is possible if we simply pass the starting and ending indices along
with the list when we make the recursive call. We leave this as an
exercise.
It is important to notice that the merge_sort function requires extra
space to hold the two halves as they are extracted with the slicing
operations. This additional space can be a critical factor if the list
is large and can make this sort problematic when working on large data
sets.
.. admonition:: Self Check
.. mchoice:: question_sort_5 :correct: b :answer_a: [16, 49, 39, 27, 43, 34, 46, 40] :answer_b: [21,1] :answer_c: [21, 1, 26, 45] :answer_d: [21] :feedback_a: This is the second half of the list. :feedback_b: Yes, mergesort will continue to recursively move toward the beginning of the list until it hits a base case. :feedback_c: Remember mergesort doesn't work on the right half of the list until the left half is completely sorted. :feedback_d: This is the list after 4 recursive calls
Given the following list of numbers: <br> [21, 1, 26, 45, 29, 28, 2, 9, 16, 49, 39, 27, 43, 34, 46, 40] <br> which answer illustrates the list to be sorted after 3 recursive calls to mergesort?
.. mchoice:: question_sort_6 :correct: c :answer_a: [21, 1] and [26, 45] :answer_b: [[1, 2, 9, 21, 26, 28, 29, 45] and [16, 27, 34, 39, 40, 43, 46, 49] :answer_c: [21] and [1] :answer_d: [9] and [16] :feedback_a: The first two lists merged will be base case lists, we have not yet reached a base case. :feedback_b: These will be the last two lists merged :feedback_c: The lists [21] and [1] are the first two base cases encountered by mergesort and will therefore be the first two lists merged. :feedback_d: Although 9 and 16 are next to each other they are in different halves of the list starting with the first split.
Given the following list of numbers: <br> [21, 1, 26, 45, 29, 28, 2, 9, 16, 49, 39, 27, 43, 34, 46, 40] <br> which answer illustrates the first two lists to be merged?
创建日期: 2023年10月10日