chore(docs): sync doc metadata [skip ci]

Author: longsizhuo

app/docs/CommunityShare/Leetcode/[1333]餐厅过滤器_translated.md

@@ -1,25 +1,28 @@
 ---
 title: 1333.Restaurant filter
-date: '2024.01.01 0:00'
+date: "2024.01.01 0:00"
 tags:
   - Python
   - answer
   - Sort
   - Array
 abbrlink: 7f1331bc
+docId: jcqhknk5z2xr3rfqn49me4j9
 ---
+
 # topic：
 
-        
 [1333.Restaurant filter.md](https://leetcode-cn.com/problems/filter-restaurants-by-vegan-friendly-price-and-distance/)
 
 # 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。
+
 ```python
 while ind < len(restaurants):
      i = restaurants[ind]
@@ -34,6 +37,7 @@ while ind < len(restaurants):
 ```
 
 # Code：
+
 ```python
 class Solution:
     def filterRestaurants(self, restaurants: List[List[int]], veganFriendly: int, maxPrice: int, maxDistance: int) -> \
@@ -46,4 +50,4 @@ class Solution:
         ]
         restaurants = sorted(restaurants, key=lambda x: (x[1], x[0]), reverse=True)
         return [i[0] for i in restaurants]
-```
+```

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