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 betweeni 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]]

在一个数组中找到满足以下条件的三元组（i,j,k）的个数

• 1.要求(i,j)的欧氏距离等于(i,k)

• 2.(i,k,j)和(j,k,i)为不同

public int distance(int a[],int b[])
    {
        int x = a[0] - b[0];
        int y = a[1] - b[1];
        return x*x + y*y;
    }
    public int numberOfBoomerangs(int[][] points) {//[a,b,c,d]
        int result = 0;
        HashMap<Integer,Integer> map = new HashMap<>();
        for(int i  = 0; i < points.length; i++)
        {
            for(int j = 0; j < points.length ; j++) //内层循环，计算ab,ac,ad距离
            {
                if(i != j)
                {
                    int dis = distance(points[i], points[j]);
                    map.put(dis, map.getOrDefault(dis, 0) + 1); //类似python a.get(key,default)
                }
            }
            for(int val:map.values()) 
                result += val * (val-1); //全排列 A(n,2)
            map.clear(); //清空map
        }
        
        return result;
    }

在一次循环中，遍历所有的点，找出任意两个点的距离，例如(ab,ac,ad)，并保存距离为d个组合个数n。由于条件2，相当于全排列A(n,2)。

知识点：

• 1.map获取值的方式map.getOrDefault(Object key, Integer defaultValue)类似与python的dict的get

• 2.map.values的用法，类似的有map.keySet()