1333. Filter Restaurants by Vegan-Friendly, Price and Distance

问题：

给出一组餐厅各种参数的，餐厅排行榜，

根据用户提供的3个要求：Vegan-Friendly， Price ，Distance

过滤满足要求的餐厅，并按照餐厅排行榜排序，若排名ranking一样，则按照id排序（都是大的排在前面）。

Example 1:
Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 1, maxPrice = 50, maxDistance = 10
Output: [3,1,5] 
Explanation: 
The restaurants are:
Restaurant 1 [id=1, rating=4, veganFriendly=1, price=40, distance=10]
Restaurant 2 [id=2, rating=8, veganFriendly=0, price=50, distance=5]
Restaurant 3 [id=3, rating=8, veganFriendly=1, price=30, distance=4]
Restaurant 4 [id=4, rating=10, veganFriendly=0, price=10, distance=3]
Restaurant 5 [id=5, rating=1, veganFriendly=1, price=15, distance=1] 
After filter restaurants with veganFriendly = 1, maxPrice = 50 and maxDistance = 10 we have restaurant 3, restaurant 1 and restaurant 5 (ordered by rating from highest to lowest). 

Example 2:
Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 50, maxDistance = 10
Output: [4,3,2,1,5]
Explanation: The restaurants are the same as in example 1, but in this case the filter veganFriendly = 0, therefore all restaurants are considered.

Example 3:
Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 30, maxDistance = 3
Output: [4,5]
 

Constraints:

1 <= restaurants.length <= 10^4
restaurants[i].length == 5
1 <= idi, ratingi, pricei, distancei <= 10^5
1 <= maxPrice, maxDistance <= 10^5
veganFriendlyi and veganFriendly are 0 or 1.
All idi are distinct.

　　

解法：

遍历餐厅排行榜，选出满足3个条件的餐厅，保存ranking和id到结果res数组

然后使用sort函数，进行降序排序

首先比较ranking，ranking降序排序，

若ranking一样，则比较id，id降序排序。

代码参考：

class Solution {
public:
    static bool comp(const vector<int>& a, const vector<int>&b){
        return (a[1]==b[1])?a[0]>b[0]:a[1]>b[1];
    }
    vector<int> filterRestaurants(vector<vector<int>>& restaurants, int veganFriendly, int maxPrice, int maxDistance) {
        vector<vector<int>> res;//{res[i][0]=id,res[i][1]=rating}
        vector<int> resid;
        for(vector<int> rst:restaurants){
            if(((veganFriendly==1 && rst[2]==1) || (veganFriendly==0)) && rst[3]<=maxPrice && rst[4]<=maxDistance){
                res.push_back({rst[0], rst[1]});
            }
        }
        sort(res.begin(), res.end(), comp);
        for(auto a:res){
            resid.push_back(a[0]);
        }
        return resid;
    }
};