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]]

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
     
       map = new HashMap
      
       (); 6             for (int j=0; j
       
         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
     
       map = new HashMap<>(); 5     for(int i=0; i