leetcode447. Number of Boomerangs

问题描述：

  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 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]]

思路

  利用for循环，逐个点进行分析其他n-1个点与其的距离。将距离和计数存放到HashMap中进行存储，如与a距离相等的点有3个，那组合形式一共有3*2共6种；若有4个，则有4*3共12种。

代码

class Solution {    public int numberOfBoomerangs(int[][] points) {        if (points == null) return 0;        int m = points.length;        int result = 0;        for(int i = 0; i < m; i++){            Map<Integer, Integer> map = new HashMap();            for(int j = 0; j < m; j++){                if(i==j) continue;                int dx = points[i][0] - points[j][0];                int dy = points[i][1] - points[j][1];                int distance = dx * dx + dy * dy;                map.put(distance,map.getOrDefault(distance,0)+1);            }            for(int k : map.keySet()){                int n = map.get(k);                result += n*(n-1);            }        }        return result;            }}

HashMap方法简介

1. map.put(K key, V value)：向集合内插入内容，key为键名，value为对应键名的键值；
2. map.getOrDefault(key, V)：在map中搜索符合提供的键的值，如果没有找到提供的键的匹配项，则返回一个默认值V；
3. map.keySet()：取出map中所有的键放入集合中；
4. map.get(key)：获取键对应的值。