leetcode406. Rebuild the queue based on height

One ： You don't have to turn against the wind , But please be born to the sun

Two ： subject

3、 ... and ： Upper code

class Solution {
    
public:
    /**  Analyze the meaning of the topic : This gives people In the array , We need according to its elements people[i] = [hi,ki]; And then according to its ki To indicate greater than hi  The number of tall people , To sort .  Ideas :1. First we have to sort , So which key value do we sort by ? 1>: If the k Value sorting , It won't work  people = [[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]  If the k If it's worth it   Then [4,4] will   At the end , It violates its front only 4 A rule that height is higher than height  2>: Then press h sorted   From high to low   If we meet people of similar height, we need to k It's worth putting the small one in front ( because k It means before   The height of the face is higher ), When the order is arranged in this way, the rest is to insert .  After arranging the order : [[7,0],[7,1],[6,1],[5,0],[5,2],[4,4]]  We take [5,2] give an example ,[5,2] The front one must be better than 5 higher , Then we can insert it in the subscript 2 The location of , Because the front   Subscript to be 0,1 Is taller than his height  */
    static bool cmp(const vector<int> &v1,const vector<int> &v2) {
    
        if(v1[0] == v2[0]) return v1[1] < v2[1];// When the body height is the same , Let's compare k value 
        return v1[0] > v2[0];
    }

    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
    
        sort(people.begin(),people.end(),cmp);
        vector<vector<int> >ans;

        for(int i = 0; i < people.size(); i++) {
    
            int postion = people[i][1];// Get the subscript to insert 
            ans.insert(ans.begin()+postion,people[i]);
        }

        return ans;
    }   
};