问题描述
我对递归合并排序的Python版本有疑问。我完成了仅由数组引用的基本版本，现在正在使用索引版本。我将陷入无尽的循环，但是我不确定我做错了什么。您愿意分享一些想法吗？预先谢谢你。
成功且无索引的版本：
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
# 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
预期结果：
x = random.sample(range(0,30),15)
mergesort(x)
[0,1,3,6,9,10,11,13,14,17,18,20,25,27,29]
但是我将陷入索引版本的无限循环：
def ms(x,left,right):
# the base case: right == left as a single-element array
if left < right:
mid = (left + right) // 2
ms(x,mid)
ms(x,mid,right + 1)
m(x,right)
return m
def m(x,right):
nlower = mid - left
nupper = right - mid + 1
temp = [0] * (nlower + nupper)
ilower,iupper,k = left,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
测试数据为：
x = random.sample(range(0,15)
ms(x,14)
---------------------------------------------------------------------------
RecursionError Traceback (most recent call last)
<ipython-input-59-39859c9eae4a> in <module>
1 x = random.sample(range(0,15)
----> 2 ms(x,14)
... last 2 frames repeated,from the frame below ...
<ipython-input-57-854342dcdefb> in ms(x,right)
3 if left < right:
4 mid = (left + right)//2
----> 5 ms(x,mid)
6 ms(x,right+1)
7 m(x,right)
RecursionError: maximum recursion depth exceeded in comparison
您介意提供一些见解吗？谢谢。
解决方法
您的索引版本使用一个令人困惑的约定，其中left是切片中第一个元素的索引，而right是最后一个元素的索引。此约定要求容易出错的+1 / -1调整。您的问题是这样的：mid的计算结果是左半部分最后一个元素的索引，但是您认为mid是右半部分的第一个元素：将2个元素切片分成一个为0，另一个为2，因此无限递归。您可以通过将ms(x,mid,right+1)更改为ms(x,mid+1,right)等来解决此问题。
此外，从功能m重新调整ms毫无意义。如果有的话，您应该返回x。
right成为最后一个元素之后的索引的可能性要小得多，就像Python中的范围说明符一样。这样，就不会出现令人困惑的+1 / -1调整。
此处是修改版本：
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,mid)
ms(x,right)
m(x,right)
return x
def m(x,right):
nlower = mid - left
nupper = right - mid
temp = [0] * (nlower + nupper)
ilower,iupper,k = left,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
您可以按以下方式调用
：x = random.sample(range(0,30),15)
ms(x,len(x))