因此,我决定通过创建另一个与要排序的输入数组(arraynumber)长度相同的数组(比如arraycount)来解决这个问题。为了让程序真正检查排序函数的稳定性,我决定使用重复出现的整数数组的测试用例,例如arraynumber = {10,8,7,8,2,2,2,1,6}
。然后我使用一个函数来计算数组中出现的数字,并将它们存储在arraycount[]
中,结果将是arraycount={1,1,1,2,1,2,3,1,1}
。
自然地,在按递增顺序排序之后,它们变成arraynumber={1,2,2,2,6,7,8,8,10}
,数组计数的结果决定了它是否稳定。因此,如果它稳定arraycount={1,1,2,3,1,1,1,2,1}
,否则就不一样了。
这是我一直在写的代码。感谢您的帮助和反馈。主要功能和输入还没有采取。我将采取另一个数组副本的原始arraycount这将是比较后,检查是否保留原来的位置。如果我犯了什么小错误,我很抱歉。非常感谢。你知道吗
def merge(arr1, l, m, r,arr2):
n1 = m - l + 1
n2 = r- m
# create temp arrays
L = [0] * (n1)
R = [0] * (n2)
# Copy data to temp arrays L[] and R[]
for i in range(0 , n1):
L[i] = arr1[l + i]
for j in range(0 , n2):
R[j] = arr1[m + 1 + j]
# Merge the temp arrays back into arr[l..r]
i = 0 # Initial index of first subarray
j = 0 # Initial index of second subarray
k = l # Initial index of merged subarray
while i < n1 and j < n2 :
if L[i] <= R[j]:
arr1[k] = L[i]
arr2[k]=L[i]
i += 1
else:
arr1[k] = R[j]
arr2[k]=R[j]
j += 1
k += 1
# Copy the remaining elements of L[], if there
# are any
while i < n1:
arr1[k] = L[i]
arr2[k]=L[i]
i += 1
k += 1
# Copy the remaining elements of R[], if there
# are any
while j < n2:
arr1[k] = R[j]
arr2[k]=R[j]
j += 1
k += 1
# l is for left index and r is right index of the
# sub-array of arr to be sorted
def mergeSort(arr1,l,r,arr2):
if l < r:
# Same as (l+r)/2, but avoids overflow for
# large l and h
m = (l+(r-1))/2
# Sort first and second halves
mergeSort(arr1, l, m,arr2)
mergeSort(arr1, m+1, r,arr2)
merge(arr1, l, m, r,arr2)
# This function takes last element as pivot, places
# the pivot element at its correct position in sorted
# array, and places all smaller (smaller than pivot)
# to left of pivot and all greater elements to right
# of pivot
def partition(arr1,low,high,arr2):
i = ( low-1 ) # index of smaller element
pivot = arr1[high] # pivot
for j in range(low , high):
# If current element is smaller than or
# equal to pivot
if arr1[j] <= pivot:
# increment index of smaller element
i = i+1
arr1[i],arr1[j] = arr1[j],arr1[i]
arr2[i],arr2[j]=arr2[j],arr2[i]
arr1[i+1],arr1[high] = arr1[high],arr1[i+1]
arr2[i+1],arr2[high]=arr2[high],arr2[i+1]
return ( i+1 )
# The main function that implements QuickSort
# arr1[] --> Array to be sorted,
# arr2[] --> Second Array of the count to be sorted along with first array
# low --> Starting index,
# high --> Ending index
# Function to do Quick sort
def quickSort(arr1,low,high,arr2):
if low < high:
# pi is partitioning index, arr[p] is now
# at right place
pi = partition(arr1,low,high,arr2)
# Separately sort elements before
# partition and after partition
quickSort(arr1, low, pi-1,arr2)
quickSort(arr1, pi+1, high,arr2)
def countArray(arr1,arr2):
i=0,j=1,k=0
while i<range(len(arr1[])):
while j<range(len(arr1[])):
if arr1[i]==arr1[j]:
arr2[i]=k+1
j=j+1
k=0
i=i+1
arraylist=[]
arraycount=[]
def isStable(arr1,arr2):
k=0
while i < range(len(arr1[])):
if arr[i]==arr2[i]:
k=k+1
i=i+1
if k==len(arr2):
print("Stable sort")
else:
print("Unstable sort")
目前没有回答
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