Solution.java

/// Source : https://leetcode.com/problems/number-of-boomerangs/description/
/// Author : liuyubobobo
/// Time   : 2017-11-15

import java.util.HashMap;

/// Using Hash Map
/// Time Complexity: O(n^2)
/// Space Complexity: O(n)
public class Solution {

    public int numberOfBoomerangs(int[][] points) {

        int res = 0;
        for( int i = 0 ; i < points.length ; i ++ ){

            // record中存储 点i 到所有其他点的距离出现的频次
            HashMap<Integer, Integer> record = new HashMap<Integer, Integer>();
            for(int j = 0 ; j < points.length ; j ++)
                if(j != i){
                    // 计算距离时不进行开根运算, 以保证精度
                    int dis = dis(points[i], points[j]);
                    if(record.containsKey(dis))
                        record.put(dis, record.get(dis) + 1);
                    else
                        record.put(dis, 1);
            }

            for(Integer dis: record.keySet())
                res += record.get(dis) * (record.get(dis) - 1);
        }

        return res;
    }

    private int dis(int[] pa, int pb[]){
        return (pa[0] - pb[0]) * (pa[0] - pb[0]) +
               (pa[1] - pb[1]) * (pa[1] - pb[1]);
    }

    public static void main(String[] args) {

        int[][] points = {{0, 0}, {1, 0}, {2, 0}};
        System.out.println((new Solution()).numberOfBoomerangs(points));
    }
}