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]]
题目含义：定义一种类似“回形标”的三元组结构，即在三元组(i, j, k)中i和j之间的距离与i和k之间的距离相等。找到一组坐标数据中，可构成几组这样的“回形标”结构。

    private int getDistance(int[] p1, int[] p2) {
        int x = p1[0] - p2[0];
        int y = p1[1] - p2[1];
        return x * x + y * y;
    }
    
    public int numberOfBoomerangs(int[][] points) {
        if (points.length == 0 || points[0].length == 0) return 0;
        int result = 0;
        for (int i = 0; i < points.length; i++) {
            Map<Integer, Integer> distances = new HashMap<>();
            for (int j = 0; j < points.length; j++) {
                if (i == j) continue;
                int distance = getDistance(points[i], points[j]);
                distances.put(distance, distances.getOrDefault(distance, 0) + 1);
            }
            for (Integer count : distances.values()) {
                result += count * (count - 1); //只有大于两条的才能有值.count个节点距离都相等，任选两条就可以与其它边一起构成会回形标。所以可以构成count * (count - 1)条
            }
        }
        return result;     
    }