447. Number of Boomerangs

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:
2

Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

本题虽然是easy题目，但是题目确实还是挺不好想的，首先我们要明白，原问题要求我们求出相同距离，而之前的题目Line reflection要求我们求对称线，Max lines on a plane要求我们求
k值相同。那么问题来了，怎么求出相同距离呢，我们这里需要用到圆的概念：圆心到圆上任意点的距离相等。接下来再处理一下boomerang的概念，即：寻找有多少对等长的线段*2；代码如下：

public class Solution {
    public int numberOfBoomerangs(int[][] points) {
        Map<Integer,Integer> map = new HashMap<>();
        int res= 0 ;
        for(int i=0;i<points.length;i++){
            for(int j=0;j<points.length;j++){
                if(i==j) continue;
                int dis = distance(points[i],points[j]);
                map.put(dis,map.getOrDefault(dis,0)+1);
            }
            int val = 0;
            for(int v:map.values()){
                val+=v*(v-1);
            }
            res+=val;
            map.clear();
        }
        return res;
    }
    public int distance(int[] a,int[] b){
        int x = a[0]-b[0];
        int y = a[1]-b[1];
        return x*x+y*y;
    }
}