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]]

给定平面上所有两两不同的n个点，“回飞棒”是一组点(i, j, k)，使i和j之间的距离等于i和k之间的距离(元组的顺序)。
找出回飞棒的数量。您可以假设n最多为500，点的坐标都在[-10000,10000](包括)范围内。

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

//新建hashmap hm
HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();

for(int count1=0;count1<points.length;count1++)
{
for(int count2=0;count2<points.length;count2++)
{
if(count1==count2)
continue;
else{
int distance=distance(points[count1],points[count2]);

//Map的新方法getOrDefault(Object,V)允许调用者在代码语句中规定获得在map中符合提供的键的值，否则在没有找到提供的键的匹配项的时候返回一个“默认值”。
hm.put(distance,hm.getOrDefault(distance,0)+1);
}
}
for(int val:hm.values())
{
result+=val*(val-1);
}
hm.clear();
}
return result;

}
int distance(int[] point1,int[] point2)
{

//返回两点间的距离
return (point1[0]-point2[0])*(point1[0]-point2[0])+(point1[1]-point2[1])*(point1[1]-point2[1]);
}
}