[Leetcode] 447. Number of Boomerangs 解题报告

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本文介绍了一种计算平面上特定几何结构——Boomerang的数量的方法。Boomerang定义为由三个点组成的元组，其中两个点到第三个点的距离相等。通过计算每一点与其他点之间的距离并统计相同距离出现的次数，最终得出Boomerang的总数。

题目：

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 iand 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：

对于每个点，计算出来它和其它所有点的距离。对于每个距离，假设有m个点，则可以构成m * (m - 1)个Boomerangs，返回所有结果之和即可。思路挺简单。

代码：

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int ret = 0;
        for(int i = 0; i < points.size(); ++i) {
            unordered_map<long, int> hash;
            for(int j = 0; j < points.size(); ++j) {
                if(j == i) {
                    continue;
                }
                long dist = (points[i].first - points[j].first) * (points[i].first - points[j].first);
                dist += (points[i].second - points[j].second) * (points[i].second - points[j].second);
                ++hash[dist];
            }
            for(auto val : hash) {
                ret += val.second * (val.second - 1);
            }
        }
        return ret;
    }
};