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;
}