Leetcode: 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]]

Solution: Use HashTable, Time: O(N^2), Space: O(N)

我的：注意14行是有value个重复distance，表示有value个点，他们跟指定点距离都是distance，需要选取2个做permutation, 所以是value * (value-1)

 public class Solution {
     public int numberOfBoomerangs(int[][] points) {
         int res = 0;
         for (int i=0; i<points.length; i++) {
             HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
             for (int j=0; j<points.length; j++) {
                 if (i == j) continue;
                 int dis = calDistance(points[i], points[j]);
                 if (!map.containsKey(dis)) map.put(dis, 1);
                 else map.put(dis, map.get(dis)+1);
             }
             for (int value : map.values()) {
                 if (value > 1) {
                     res += value * (value - 1);
                 }
             }
         }
         return res;
     }

     public int calDistance(int[] p1, int[] p2) {
         int dx = Math.abs(p2[0] - p1[0]);
         int dy = Math.abs(p2[1] - p1[1]);
         return dx*dx + dy*dy;
     }
 }

别人的简洁写法

 public int numberOfBoomerangs(int[][] points) {
     int res = 0;

     Map<Integer, Integer> map = new HashMap<>();
     for(int i=0; i<points.length; i++) {
         for(int j=0; j<points.length; j++) {
             if(i == j)
                 continue;

             int d = getDistance(points[i], points[j]);
             map.put(d, map.getOrDefault(d, 0) + 1);
         }

         for(int val : map.values()) {
             res += val * (val-1);
         }
         map.clear();
     }

     return res;
 }

 private int getDistance(int[] a, int[] b) {
     int dx = a[0] - b[0];
     int dy = a[1] - b[1];

     return dx*dx + dy*dy;
 }