1.问题描述

描述：

　　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 jequals 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).

示例：

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]]

2.问题分析

1. 第一遍：
   • 如果到点a的距离为distance的点共有n个，那么从这n个点中有序找两个点就能与A点构成回旋，根据排列公式，总的回旋个数为A_n²个，即n * (n - 1)个。遍历每一个点，map的key是距离，value是距离该点key个单位的点的个数。

     class Solution {
         public int numberOfBoomerangs(int[][] points) {
             int res = 0;
             for(int i = 0; i < points.length; i++){
                 Map<Double, Integer> map = new HashMap<>();
                 for(int j = 0; j < points.length; j++){
                     Double distance = Math.pow(points[i][0] - points[j][0], 2) + Math.pow(points[i][1] - points[j][1], 2);
                     if(map.containsKey(distance)){
                         map.put(distance, map.get(distance) + 1);
                     }else{
                         map.put(distance, 1);
                     }
                 }
                 
                 for(Integer count : map.values()){
                     res += count * (count - 1);
                 }
             }
             return res;
         }
     }

     

   • 改进：将map移到for循坏外，之后每次外部循环结束后，执行map.clear()操作。不再每次都新建map集合。
     

             Map<Double, Integer> map = new HashMap<>();
             for(int i = 0; i < points.length; i++){
                 for(int j = 0; j < points.length; j++){
                     Double distance = Math.pow(points[i][0] - points[j][0], 2) + Math.pow(points[i][1] - points[j][1], 2);
                     if(map.containsKey(distance)){
                         map.put(distance, map.get(distance) + 1);
                     }else{
                         map.put(distance, 1);
                     }
                 }
                 
                 for(Integer count : map.values()){
                     res += count * (count - 1);
                 }
                 map.clear();
             }

     View Code