[Leetcode] 447. Number of Boomerangs 解题报告
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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 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]]
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; }};
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