Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between iand j equals the distance between i and k (the order of the tuple matters).

Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000] (inclusive).

Example:

Input:[[0,0],[1,0],[2,0]]Output:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]] 给3个点，计算1个点到另外两个点都相等有多少种组合 hypot(x,y)   计算 x 与 y 平方和的平方根 C++(386ms):

1 class Solution { 2 public: 3     int numberOfBoomerangs(vector
     
      
       >& points) { 4         int res = 0; 5         unordered_map
       
         ctr(points.size()); 6         for (auto p : points) { 7             for (auto q : points){ 8                 if (p == q) 9                     continue ;10                 double t = hypot(p.first - q.first, p.second - q.second) ;11                 ctr[t]++ ;12                 res += 2 * (ctr[t]-1);13             }14             ctr.clear() ;15         }16         return res;17     }18 };