1333. Filter Restaurants by Vegan-Friendly, Price, and Distance
Problem
1333. Filter Restaurants by Vegan-Friendly, Price, and Distance
Approach
My initial approach used pop(), but the time complexity was too high (~400ms). I switched to a list comprehension.
A senior once told me: pop() runtime is fine, but when you combine it with index operations inside, it becomes very slow.
sorted and lambda usage:
lambda is Python's anonymous function syntax. Here, lambda x: (x[1], x[0]) defines a function that takes an element x (a sublist of restaurants) and returns a tuple (x[1], x[0]).
This means sorting is first based on x[1] (rating), then by x[0] (id) if x[1] values are equal.
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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