LeetCode 447. Number of boomerangs (permutation and combination problem)

In general, we need to find the combination of three points ,i,j,k, For i Come on ,j and k On i The distances must be equal .
Violent enumeration requires a triple cycle , So the time complexity is O(n^3) It's not worth it .
So we can think of the permutation and combination problem , The idea is to enumerate if i Fixed point , Filter out the distance and... From the remaining points i A collection of equal points .
Suppose you're away from i A distance of a A little n individual , So it's from n Choose two points in the order and i Combine , The permutation and combination problem ,An2.
Store a hash table with a distance of a Number of combinations , Add it up .

int numberOfBoomerangs(vector<vector<int>>& points) 
    {
    
        int res=0;

        for(int i=0;i<points.size();i++)
        {
    
            unordered_map<int,int> map;
            for(int j=0;j<points.size();j++)
            {
    
                int dis=Distance(points[i],points[j]);
                map[dis]++;
            }
            for(auto p=map.begin();p!=map.end();p++)
            {
    
                res+=p->second*(p->second-1);
            }
        }
        return res;
    }
    
    int Distance(vector<int> a, vector<int> b)
    {
    
        int x=a[0]-b[0];
        int y=a[1]-b[1];
        
        return x*x+y*y;
    }