447. Number of Boomerangs(飞去来器)

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]]

方法一：

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int count=0;
        for(auto i:points)
        {
            unordered_map<long,int> distance;
            for(auto j:points)
                if(i!=j)
                    distance[pow(i.first-j.first,2)+pow(i.second-j.second,2)]++;
            for(auto k:distance)
                if(k.second>1)
                    count +=k.second*(k.second-1);//排列组合
        }
        return count;
    }
};

方法二：摘自讨论区

class Solution{
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int booms = 0;
        for (auto &p : points) {
            unordered_map<double, int> ctr(points.size());
            for (auto &q : points)
                booms += 2 * ctr[hypot(p.first - q.first, p.second - q.second)]++;//每增加一个可以凑成同样距离的 “boomerang” 点之后，我们其实就要增加 2 * 当前同距离点的个数 个 “boomerang” 元组可能（也就是该点到同距离的各个点之后，再换个位置，所以 * 2）
        }
        return booms;
    }
};