LeetCode-Number of Boomerangs

Description：
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).

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]]

题意：给定一系列的坐标点，找出所有可能的点$(i, j, k)$，使得满足$dis(i, j) == dis(i, k)$；

解法：我们需要考虑每一个点，计算是否有另外的两个点满足到当前点的距离相同；

1. 遍历所有的点$i$；
2. 遍历除$i$外的所有点$j$，利用散列表$table$记录下此时点$j$到点$i$的距离，并且对于每个点$j$，在散列表中查找是否有其他的点到$i$的距离相同；则此时可能的坐标点组合为$2*table(dis(i, j))$，每包含一个具有相同距离的点，其组合数增加2（即$(i,j,k)$和$(i,k,j)$）；

Java

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int result = 0;
        for (int[] i : points) {
            Map<Integer, Integer> table = new HashMap<>();
            for (int[] j : points) {
                if (i == j) continue;
                int dis = (i[0] - j[0]) * (i[0] - j[0]) + 
                    (i[1] - j[1]) * (i[1] - j[1]);
                int preDis = table.getOrDefault(dis, 0);
                result += 2 * preDis;
                table.put(dis, preDis + 1);
            }
        }
        return result;
    }
}