[LeetCode]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]]

解法：
双重循环，记录每个点和其他的点的距离。

public int numberOfBoomerangs(int[][] points) {
        int result = 0;
        for(int i = 0; i< points.length; i++){
            Map<Integer,Integer> hash = new HashMap<>();
            for( int j = 0; j< points.length; j++){
                if(i == j){
                    continue;
                }else{
                    int dist = getDistance(points[i],points[j]);
                    hash.put(dist,hash.getOrDefault(dist,0)+1);
                }
            }
            for(Integer val : hash.values()){
                result += val * (val - 1);
            }
        }
        return result;
    }

    int getDistance(int[] p1, int[] p2){
        int x = p1[0] - p2[0];
        int y = p1[1] - p2[1];
        return x*x + y*y;
    }