447. Number of Boomerangs 

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class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int count = 0;

        for (int i = 0; i < points.length; i++) {
            // 计算出与其他点的距离
            Map<Integer, Integer> powerToCountMap = new HashMap<>();

            // 以当前点为中点，枚举另一个点
            for (int j = 0; j < points.length; j++) {
                if (i == j) {
                    continue;
                }
                int xDistance = points[j][0] - points[i][0], yDistance = points[j][1] - points[i][1];
                int powerDistance = xDistance * xDistance + yDistance * yDistance;

                int sameDistanceCount = powerToCountMap.getOrDefault(powerDistance, 0);
                count += sameDistanceCount * 2; // 可以组成回旋镖，则累加两倍元组
                powerToCountMap.put(powerDistance, sameDistanceCount + 1);
            }
        }

        return count;
    }
}

以为有什么优于 O(n^2) 的解法，结果并没有。

References

447. Number of Boomerangs