题目描述:

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:

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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]]

这道题似乎还没有什么好解法, 现在Runtime1368ms都可以beats 100%的C++提交…

现在已经不是了, 1000多ms的Runtime算是比较慢啦, 但是我暂时还没有时间来搞这道题…

使用双重循环+哈希表. 如果认为unordered_map的查询复杂度是O(1), 总体复杂度是O(n^2).

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class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int ans = 0;
        for(int i = 0; i < points.size(); i++){
            unordered_map<long long, int> distance;
            for(int j = 0; j < points.size(); j++){
                if(j == i) continue;
                long long d = getDistance(points[i], points[j]);
                if(distance.count(d)){
                    ans += 2 * distance[d];
                }
                distance[d]++;
            }
        }
        return ans;
    }
    
    long long getDistance(pair<int, int> &p1, pair<int, int> &p2){
        return (long long)(p1.first - p2.first) * (long long)(p1.first - p2.first) 
             + (long long)(p1.second - p2.second) * (long long)(p1.second - p2.second);
    }
};