LeetCode-447. Number of Boomerangs (Java)

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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

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题意

求与某一点距离相同的坐标个数

思路

循环数组，计算距离，保存距离，然后统计个数

代码

public class Solution {    public int numberOfBoomerangs(int[][] points) {       Map<Integer,Integer> map = new HashMap<Integer,Integer>();       int res =0;       for(int i=0;i<points.length;i++){   for(int j=0;j<points.length;j++){if(i==j) continue;List<Integer> list = new ArrayList<Integer>();int distance = getDistance(points[i],points[j]);if(map.containsKey(distance)){map.put(distance,map.get(distance)+1);}else{map.put(distance,1);}   }   for(int value : map.values()){res += value * (value -1);   }map.clear();        }        return res;     }    public static int getDistance(int[] pointOne,int[] pointTwo){    int distance = (int)Math.pow(pointOne[0]-pointTwo[0], 2) + (int)Math.pow(pointOne[1]-pointTwo[1], 2);    return distance;    }}