标签：bsp oom order ++ get 解法 其他 tin tor
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 iand 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]]
解法：
双重循环，记录每个点和其他的点的距离。
public int numberOfBoomerangs(int[][] points) { int result = 0; for(int i = 0; i< points.length; i++){ Map<Integer,Integer> hash = new HashMap<>(); for( int j = 0; j< points.length; j++){ if(i == j){ continue; }else{ int dist = getDistance(points[i],points[j]); hash.put(dist,hash.getOrDefault(dist,0)+1); } } for(Integer val : hash.values()){ result += val * (val - 1); } } return result; } int getDistance(int[] p1, int[] p2){ int x = p1[0] - p2[0]; int y = p1[1] - p2[1]; return x*x + y*y; }
[LeetCode]447 Number of Boomerangs
标签：bsp oom order ++ get 解法 其他 tin tor
原文地址：http://www.cnblogs.com/javanerd/p/6058195.html
