LEETCODE--Number of Boomerangs

最新推荐文章于 2019-03-24 14:57:04 发布

prettysky123

最新推荐文章于 2019-03-24 14:57:04 发布

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本文介绍了一种算法，用于计算平面上给定点集中特定类型的三元组（回旋镖）的数量。这些回旋镖是由三个点组成，其中两个点到第三个点的距离相等，且顺序重要。通过使用C++中的map来分组具有相同距离的点对，该算法有效地解决了问题。

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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]]

熟悉使用map

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int num = 0;
        for(int i = 0; i < points.size(); i++){
             map<long,int> group;
             for(int j = 0; j < points.size(); j++){
                 if(i == j)
                    continue;
                 int distance_x = points[j].first - points[i].first;
                 int distance_y = points[j].second - points[i].second;
                 int key = distance_x * distance_x + distance_y * distance_y;
                 ++group[key];
             }
             for(auto& p : group){
                 int val = p.second;
                 if(val > 1)
                    num += val * (val - 1); 
             } 
        }
        return num;

    }
};