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]]遍历每个point，然后找和它等距离的其他point，按距离来保存，比如有一个点a，和它距离都是1的点有b，c，d，那么一共的组合就有6种，包括：[a, b, c], [a, c, b], [a, b, d], [a, d, b], [a, c, d], [a, d, c]。这么算是不考重复的情况下。还有可能b, c坐标完全相同，那么b和c会被当成两个点算两次。
public class Solution { public int numberOfBoomerangs(int[][] points) { int result = 0; for(int i = 0; i < points.length; i++) { Map<Integer, Integer> map = new HashMap(); for(int j = 0; j < points.length; j++) { if(i == j) continue; int dx = points[j][0] - points[i][0], dy = points[j][1] - points[i][1]; int distance = dx * dx + dy * dy; map.put(distance, map.getOrDefault(distance, 0) + 1); } for(int k : map.keySet()) { int n = map.get(k); result += n * (n - 1); } } return result; } } /** * @param {number[][]} points * @return {number} */ var numberOfBoomerangs = function(points) { const len = points.length if (len < 3) { return 0 } let count = 0 for (let i = 0; i < len; i++) { const point1 = points[i] // 用一个map存储其它点距离当前点point1有多少种距离可能 const map = new Map() for (let j = 0; j < len; j++) { if (i != j) { const point2 = points[j] const distance = getDistance(point1, point2) if (!map.has(distance)) { map.set(distance, 1) } else { map.set(distance, map.get(distance) + 1) } } } map.forEach(value => { if (value > 1) { const num = getArrangementNum(value) count += num } }) } return count }; // 计算两点之间距离，这里其实没必要开根号 function getDistance (p1, p2) { const distance = Math.sqrt(Math.pow(p1[0] - p2[0], 2) + Math.pow(p1[1] - p2[1], 2)) return distance } // 计算排列数 function getArrangementNum (n) { return n * (n - 1) }