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)

1 public class Solution {
2     public int numberOfBoomerangs(int[][] points) {
3         int res = 0;
4         for (int i=0; i<points.length; i++) {
5             HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
6             for (int j=0; j<points.length; j++) {
7                 if (i == j) continue;
8                 int dis = calDistance(points[i], points[j]);
9                 if (!map.containsKey(dis)) map.put(dis, 1);
10                 else map.put(dis, map.get(dis)+1);
11             }
12             for (int value : map.values()) {
13                 if (value > 1) {
14                     res += value * (value - 1);
15                 }
16             }
17         }
18         return res;
19     }
20
21     public int calDistance(int[] p1, int[] p2) {
22         int dx = Math.abs(p2[0] - p1[0]);
23         int dy = Math.abs(p2[1] - p1[1]);
24         return dx*dx + dy*dy;
25     }
26 }

别人的简洁写法

1 public int numberOfBoomerangs(int[][] points) {
2     int res = 0;
3
4     Map<Integer, Integer> map = new HashMap<>();
5     for(int i=0; i<points.length; i++) {
6         for(int j=0; j<points.length; j++) {
7             if(i == j)
8                 continue;
9
10             int d = getDistance(points[i], points[j]);
11             map.put(d, map.getOrDefault(d, 0) + 1);
12         }
13
14         for(int val : map.values()) {
15             res += val * (val-1);
16         }
17         map.clear();
18     }
19
20     return res;
21 }
22
23 private int getDistance(int[] a, int[] b) {
24     int dx = a[0] - b[0];
25     int dy = a[1] - b[1];
26
27     return dx*dx + dy*dy;
28 }