1333.Restaurant filter
topic：
Thought：
This is the beginningpopThought，But time complexity is too high，400ms，It is later changed to a list derivative。 I think of the senior said，pop()The running time is okay，But once the index is added inside，It will be particularly slow。 sorted and lambda Usage： lambdayesPythonAnonymous function in。it's here，lambda x: (x[1], x[0])Definitions a acceptance of an elementx（in this case，xyesrestaurantsA list in the list）And return a tuple(x[1], x[0])The function。
this means，Sort首先基于每个子列表的第二个元素x[1]，Then based on this basisx[0]。in other words，It first followsx[1]进行Sort，ifx[1]same，According tox[0]进行Sort。
while ind < len(restaurants):
i = restaurants[ind]
if veganFriendly == 1 and i[2] == 0:
restaurants.pop(ind)
elif maxPrice < i[3]:
restaurants.pop(ind)
elif maxDistance < i[4]:
restaurants.pop(ind)
else:
ind += 1Code：
class Solution:
def filterRestaurants(self, restaurants: List[List[int]], veganFriendly: int, maxPrice: int, maxDistance: int) -> \
List[int]:
restaurants = [
i for i in restaurants
if (veganFriendly == 0 or i[2] == veganFriendly)
and i[3] <= maxPrice
and i[4] <= maxDistance
]
restaurants = sorted(restaurants, key=lambda x: (x[1], x[0]), reverse=True)
return [i[0] for i in restaurants]贡献者
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