chore: add daily leetcode post [1333]餐厅过滤器_translated

Author: longsizhuo

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

@@ -0,0 +1,49 @@
+---
+title: 1333.Restaurant filter
+date: '2024.01.01 0:00'
+tags:
+  - Python
+  - answer
+  - Sort
+  - Array
+abbrlink: 7f1331bc
+---
+# 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]
+     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 += 1
+```
+
+# Code：
+```python
+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]
+```