447. Number of Boomerangs

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本文介绍了一种高效算法，用于找出平面上给定点集中所有符合特定距离条件的三元组，即所谓的“回旋镖”。该算法使用HashMap来记录每个点到其它点的距离及出现次数，最终统计出所有可能的回旋镖组合。

题目描述：

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).

思路一：

对每个点，有个HashMap，key为距离，value为点数。boomerangs的个数为value*(value-1)。

时间复杂度为O(N^2)。

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int res = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < points.length; i++)
        {
            for (int j = 0; j < points.length; j++)
            {
                if (i == j) continue;
                int dis = getDistance(points[i], points[j]);
                map.put(dis, map.getOrDefault(dis, 0) + 1);
            }
            for (int num : map.values())
                res += num * (num - 1);
            map.clear();
        }
        return res;
    }
    private int getDistance(int[] num1, int[] num2)
    {
        return (num1[0] - num2[0]) * (num1[0] - num2[0]) + (num1[1] - num2[1]) * (num1[1] - num2[1]);
    }
}