LeetCode 447. Number of Boomerangs

最新推荐文章于 2025-10-12 09:15:28 发布

原创最新推荐文章于 2025-10-12 09:15:28 发布 · 224 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

350 篇文章

订阅专栏

本文介绍了一种算法，用于计算平面上给定点集中所有回旋镖点对的数量。回旋镖点对定义为由三个点(i, j, k)组成的元组，其中点i到点j的距离等于点i到点k的距离。文章详细解释了如何通过遍历点集并使用哈希映射来高效地计算这些回旋镖点对。

447. 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]]

分析：map(m[i] = j)用来存储点与点之间距离（其实是距离的平方，不影响结果）为i的个数j。对于某一点，距离它为i的个数若为j个，则能够构成的i，j，k这样的组数是j * (j - 1)个。每次累计得到的总结果即为所求～

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int cnt = 0;
        for(int i = 0; i < points.size(); i++) {
            map<int, int> m;
            int x1 = points[i].first, y1 = points[i].second;
            for(int j = 0; j < points.size(); j++) {
                if(j == i) continue;
                int x2 = points[j].first, y2 = points[j].second;
                int dis = (x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2);
                m[dis]++;
            }
            for(auto it = m.begin(); it != m.end(); it++) {
                cnt += it->second * (it->second - 1);
            }
        }
        return cnt;
    }
};