[leetcode] 447. Number of Boomerangs

Question:

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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

Solution:

主要的思路是对每一个点，用map记录它与其他点的距离，和这个距离出现的次数，如果这个距离没有出现过，就加进map中，初始化出现次数为1；若出现过，则在返回结果中加上这个出现过的次数乘2，因为(i, j)和(i, k)相等就意味着(i, k)和(i, j)相等，然后对这个次数加1，表示把这个距离加进map中。

class Solution {public:    int numberOfBoomerangs(vector<pair<int, int>>& points) {        int ret = 0;        for (int i = 0; i < points.size(); i++) {            map<double, int> m;            for (int j = 0; j < points.size(); j++) {                if (i == j) continue;                int dis = (points[i].first - points[j].first) * (points[i].first - points[j].first)                        + (points[i].second - points[j].second) * (points[i].second - points[j].second);                if (m.find(dis) == m.end()) {                    m[dis] = 1;                } else {                    ret += m[dis] * 2;                    m[dis]++;                }            }        }        return ret;    }};