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LEETCODE--Number of Boomerangs

2019-11-11 04:16:18
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來(lái)源:轉(zhuǎn)載
供稿:網(wǎng)友

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; }};
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