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 i and 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]]

分析：循环查找每个点， 每个点都用一个hashmap存放其他点到该点得值 作为key，i，j ＝ i， k。有重复则value＋1；

计算总共有几对是用组合 n ＊ n－1；

public class Solution {    public int numberOfBoomerangs(int[][] points) {        int sum = 0;        if(points == null || points.length == 0) return sum;                for(int i = 0 ; i < points.length ; i ++){            HashMap
     
       map = new HashMap<>();            for(int j = 0; j < points.length; j++){                int dis = getDis(points[i], points[j]);                map.put(dis, map.getOrDefault(dis,0)+1);            }                for(int value : map.values()){            sum += value * (value - 1);   //2 combination;        }        map.clear();        }        return sum;    }        public int getDis(int[] a, int[] b){        return (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]);    }    }