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]]

思路 點之間距離相同，從中選3個，一個排列組合問題。 可以放入hash，map中儲存，距離相同的點的數目，然後計算。

code:

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int count=0;
        int size=points.size();
        for(auto p:points){
            unordered_map<double,int>m;
            for(auto p1:points){
                m[hypot(p.first-p1.first,p.second-p1.second)]++;
            }
            for(auto v:m){
                if(v.second>1){
                    count+=v.second*(v.second-1);
                }
            }
        }
        return count;
    }
};
 

三角函式 hypot 使用 auto遍歷