[英]Python recursive Merge Sort by index
我有一個關於遞歸合並排序的 Python 版本的問題。 我完成了僅由數組引用的基本版本,現在正在研究索引版本。 我會陷入無限循環,但我不確定我哪里做錯了。 你介意分享一些想法嗎? 先感謝您。
成功和非索引版本:
def mergesort(x):
# The base case is when the array contains less than 1 observation.
length = len(x)
if length < 2:
return x
else:
# Recursive case:merge sort on the lower subarray, and the upper subarray.
mid = (length) // 2
lower = mergesort(x[:mid])
upper = mergesort(x[mid:])
# merge two subarrays.
x_sorted = merge(lower, upper)
return (x_sorted)
def merge(lower, upper):
nlower = len(lower)
nupper = len(upper)
i, j, k = 0, 0, 0
# create a temp array to store the sorted results
temp = [0] * (nlower + nupper)
# as the lower and upper are sorted, since the base case is the single observation.
# now we compare the smallest element in each sorted array, and store the smaller one to the temp array
# repeat this process until one array is completed moved to the temp array
# store the other array to the end of the temp array
while i < nlower and j < nupper:
if lower[i] <= upper[j]:
temp[k] = lower[i]
i += 1
k += 1
else:
temp[k] = upper[j]
j += 1
k += 1
if i == nlower:
temp[i+j:] = upper[j:]
else:
temp[i+j:] = lower[i:]
return temp
預期結果:
x = random.sample(range(0, 30), 15)
mergesort(x)
[0, 1, 3, 6, 9, 10, 11, 13, 14, 17, 18, 20, 25, 27, 29]
但是我會陷入使用索引版本的無限循環:
def ms(x, left, right):
# the base case: right == left as a single-element array
if left < right:
mid = (left + right) // 2
ms(x, left, mid)
ms(x, mid, right + 1)
m(x, left, mid, right)
return m
def m(x, left, mid, right):
nlower = mid - left
nupper = right - mid + 1
temp = [0] * (nlower + nupper)
ilower, iupper, k = left, mid, 0
while ilower < mid and iupper < right + 1:
if x[ilower] <= x[iupper]:
temp[k] = x[ilower]
ilower += 1
k += 1
else:
temp[k] = x[iupper]
iupper += 1
k += 1
if ilower == mid:
temp[k:] = x[iupper:right+1]
else:
temp[k:] = x[ilower:mid]
x[left:right+1] = temp
return x
測試數據為:
x = random.sample(range(0, 30), 15)
ms(x, 0, 14)
---------------------------------------------------------------------------
RecursionError Traceback (most recent call last)
<ipython-input-59-39859c9eae4a> in <module>
1 x = random.sample(range(0, 30), 15)
----> 2 ms(x, 0, 14)
... last 2 frames repeated, from the frame below ...
<ipython-input-57-854342dcdefb> in ms(x, left, right)
3 if left < right:
4 mid = (left + right)//2
----> 5 ms(x, left, mid)
6 ms(x, mid, right+1)
7 m(x, left, mid, right)
RecursionError: maximum recursion depth exceeded in comparison
你介意提供一些見解嗎? 謝謝。
您的索引版本使用了一種令人困惑的約定,即left
是切片中第一個元素的索引, right
是最后一個元素的索引。 此約定需要容易出錯的+1
/ -1
調整。 你的問題是這樣的: mid
as計算是左半部分最后一個元素的索引,但你認為mid
是右半部分的第一個元素:2個元素的切片被分成一個0和一個2,因此無限遞歸。 您可以通過將ms(x, mid, right+1)
更改為ms(x, mid+1, right)
等來解決此問題。
此外,從函數ms
重新調整m
沒有意義。 如果有的話,您應該返回x
。
就像 Python 中的范圍說明符一樣,作為最后一個元素之后的索引的right
更不容易出錯。 這樣就不會出現令人困惑的+1
/ -1
調整。
這是修改后的版本:
def ms(x, left, right):
# the base case: right - left as a single-element array
if right - left >= 2:
mid = (left + right) // 2 # index of the first element of the right half
ms(x, left, mid)
ms(x, mid, right)
m(x, left, mid, right)
return x
def m(x, left, mid, right):
nlower = mid - left
nupper = right - mid
temp = [0] * (nlower + nupper)
ilower, iupper, k = left, mid, 0
while ilower < mid and iupper < right:
if x[ilower] <= x[iupper]:
temp[k] = x[ilower]
ilower += 1
k += 1
else:
temp[k] = x[iupper]
iupper += 1
k += 1
if ilower == mid:
temp[k:] = x[iupper:right]
else:
temp[k:] = x[ilower:mid]
x[left:right] = temp
return x
您將調用為:
x = random.sample(range(0, 30), 15)
ms(x, 0, len(x))
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