README_EN.md

447. Number of Boomerangs

中文文档

Description

You are given n points in the plane that are all distinct, where points[i] = [xi, yi]. 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).

Return the number of boomerangs.

 

Example 1:

Input: points = [[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]].

Example 2:

Input: points = [[1,1],[2,2],[3,3]]
Output: 2

Example 3:

Input: points = [[1,1]]
Output: 0

 

Constraints:

• n == points.length
• 1 <= n <= 500
• points[i].length == 2
• -104 <= xi, yi <= 104
• All the points are unique.

Solutions

Python3

class Solution:
    def numberOfBoomerangs(self, points: List[List[int]]) -> int:
        ans = 0
        for p in points:
            counter = Counter()
            for q in points:
                distance = (p[0] - q[0]) * (p[0] - q[0]) + (p[1] - q[1]) * (p[1] - q[1])
                counter[distance] += 1
            ans += sum([val * (val - 1) for val in counter.values()])
        return ans

Java

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int ans = 0;
        for (int[] p : points) {
            Map<Integer, Integer> counter = new HashMap<>();
            for (int[] q : points) {
                int distance = (p[0] - q[0]) * (p[0] - q[0]) + (p[1] - q[1]) * (p[1] - q[1]);
                counter.put(distance, counter.getOrDefault(distance, 0) + 1);
            }
            for (int val : counter.values()) {
                ans += val * (val - 1);
            }
        }
        return ans;
    }
}

TypeScript

function numberOfBoomerangs(points: number[][]): number {
    let ans = 0;
    for (let p1 of points) {
        let hashMap: Map<number, number> = new Map();
        for (let p2 of points) {
            const distance = (p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2;
            hashMap.set(distance, (hashMap.get(distance) || 0) + 1);
        }
        for (let [, v] of [...hashMap]) {
            ans += v * (v - 1);
        }
    }
    return ans;
}

Go

func numberOfBoomerangs(points [][]int) int {
	ans := 0
	for _, p := range points {
		cnt := make(map[int]int)
		for _, q := range points {
			cnt[(p[0]-q[0])*(p[0]-q[0])+(p[1]-q[1])*(p[1]-q[1])]++
		}
		for _, v := range cnt {
			ans += v * (v - 1)
		}
	}
	return ans
}

C++

class Solution {
public:
    int numberOfBoomerangs(vector<vector<int>>& points) {
        int ans = 0;
        for (const auto& p : points) {
            unordered_map<int, int> cnt;
            for (const auto& q : points) {
                ++cnt[(p[0] - q[0]) * (p[0] - q[0]) + (p[1] - q[1]) * (p[1] - q[1])];
            }
            for (const auto& [_, v] : cnt) {
                ans += v * (v - 1);
            }
        }
        return ans;
    }
};

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