447. Number of Boomerangs

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]]

 

 Approach #1: C++. Using triple cycle .[Time Limit Exceeded]

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int size = points.size();
        int counter = 0;
        for (int i = 0; i < size; ++i) {
            for (int j = 0; j < size; ++j) {
                if (i == j) continue;
                for (int k = 0; k < size; ++k) {
                    if (i == k || j == k) continue;
                    int x1 = abs(points[i].first - points[j].first);
                    int x2 = abs(points[i].first - points[k].first);
                    int y1 = abs(points[i].second - points[j].second);
                    int y2 = abs(points[i].second - points[k].second);
                    double fs = sqrt(pow(x1, 2) + pow(y1, 2));
                    double ft = sqrt(pow(x2, 2) + pow(y2, 2));
                    if (fs == ft) counter++;
                    
                }
            }
        }
        return counter;
    }
};

　　

Approach #2: C++.

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int size = points.size();
        int counter = 0;
        
        for (int i = 0; i < size; ++i) {
            unordered_map<int, int> distances;
            for (int j = 0; j < size; ++j) {
                if (i == j) continue;
                int distance = (points[i].first - points[j].first) * (points[i].first - points[j].first) + (points[i].second -                                       points[j].second) * (points[i].second - points[j].second);
                distances[distance]++;
            }
            for (auto& p : distances) {
                if (p.second > 1)
                    counter += p.second * (p.second - 1);
            }
        }

        return counter;
    }
};

　　

Approach #2: Java.

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int res = 0;
        for (int i = 0; i < points.length; ++i) {
            HashMap<Integer, Integer> map = new HashMap<>();
            for (int j = 0; j < points.length; ++j) {
                if (points[i] == points[j]) continue;
                int dx = points[i][0] - points[j][0];
                int dy = points[i][1] - points[j][1];
                int d = dx * dx + dy * dy;
                map.put(d, map.getOrDefault(d, 0) + 1);
            }
            for (int val : map.values()) {
                res += val * (val - 1);
            }
        }
        return res;
    }
}

　　

Approach #3: Python.

class Solution(object):
    def numberOfBoomerangs(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        res = 0
        for p in points:
            cmap = {}
            for q in points:
                f = p[0] - q[0]
                s = p[1] - q[1]
                cmap[f*f + s*s] = 1 + cmap.get(f*f + s*s, 0)
            for k in cmap:
                res += cmap[k] * (cmap[k] - 1)
        return res

　　

Time Submitted	Status	Runtime	Language
a few seconds ago	Accepted	241 ms	java
5 minutes ago	Accepted	1436 ms	python
8 minutes ago	Accepted	196 ms	cpp

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/9972043.html