leetcode 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 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]);
	}
}