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ilker senturk
17,583 Pointsquick sort in python
when run the method, it seems work but if we rerun loosing an element in the list,
def quickSort(sequence):
length = len(sequence)
if(length <= 1):
return sequence
else:
pivotPoint = sequence.pop()
items_greater = []
items_lower = []
for item in sequence:
if item > pivotPoint :
items_greater.append(item)
else:
items_lower.append(item)
return quickSort(items_lower) + [pivotPoint] + quickSort(items_greater)
2 Answers
Chris Freeman
Treehouse Moderator 68,441 PointsThe issue is caused by the pop() in
pivotPoint = sequence.pop()
Python passes arguments by reference, so at the top level of recursion, the first pop actually pops an item off of the arr collection.
Because of this feature of python, it is best not to directly alter arguments unless explicitly understanding the side effects. Local assignment to a variable, make it "local" (a distinct copy of the argument). For example:
def somefunc(arr):
first = arr[0]
arr = arr[1:] # arr is now local and external arr is unaffected
Another simple fix would be to use:
pivotPoint = sequence[0] # get the first item
and
for item in sequence[1:]: # sort excluding the first item
Post back if you need more help.
ilker senturk
17,583 Pointswhen i call the method , it sorted the sequence , so the output for arr = [8, 7, 2, 5, 3, 1,12] will be [1, 2, 3, 5, 7, 8, 12] but if Call the method twice
print(quickSort(arr))
print(quickSort(arr))
printing [1, 2, 3, 5, 7, 8] where i am loosing 12,
print(quickSort(arr))
print(quickSort(arr))
print(quickSort(arr))
printing [2, 3, 5, 7, 8] where i lost element 1 I couldn't get where i make a mistake
Chris Freeman
Treehouse Moderator 68,441 PointsChris Freeman
Treehouse Moderator 68,441 PointsCan you please expand on what you mean by "if we rerun loosing an element in the list"?