Elso hazi #14
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Elso hazi #14
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szabadkai
reviewed
Mar 19, 2018
| if tup[-1] == ending: | ||
| res.append(tup) | ||
| return res | ||
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wow :) Klassz megoldas, bar egy kicsit lassu. https://wiki.python.org/moin/HowTo/Sorting
key functions alatt talasz elegansabb megoldast.
szabadkai
reviewed
Mar 19, 2018
basic/list2.py
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| @@ -13,17 +13,15 @@ | |||
| # so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or | |||
| # modify the passed in list. | |||
| def remove_adjacent(nums): | |||
| # +++your code here+++ | |||
| return | |||
| return list(set(nums)) | |||
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haat... nem egeszen. mit gondolsz jo eredmenyt kapnal mondjuk [1,2,3,3,1] listara?
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Ja igen elnéztem, azt hittem csak simán halmazként kell visszaadni. Javítom majd:)
szabadkai
reviewed
Mar 19, 2018
basic/list2.py
Outdated
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| # E. Given two lists sorted in increasing order, create and return a merged | ||
| # list of all the elements in sorted order. You may modify the passed in lists. | ||
| # Ideally, the solution should work in "linear" time, making a single | ||
| # pass of both lists. | ||
| def linear_merge(list1, list2): | ||
| # +++your code here+++ | ||
| return | ||
| return sorted(list1 + list2) |
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nem rossz probalkozas, de a python sorted TimSort algoritmussal dolgozik nlogn idoben. A feladat leirasaban a megoldas; a ket lista eleve rendezett. ;)
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