447. Number of Boomerangs*-CFANZ编程社区

447. Number of Boomerangs*

https://leetcode.com/problems/number-of-boomerangs/description/

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 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]]

C++ 实现 1

给定一个 vector, 里面包含 n 个点, 如果 3 个点(索引分别为 i, j, k)满足 dis(p[i], p[j]) 等于 dis(p[i], p[k]), 那么就称 (i, j, k) 为一个 boomerang. 注意 tuple 中的元素的顺序. 现在要统计 boomerang 的个数.

思路: 由于 tuple 中有 3 个元素, 可以考虑先固定索引 i, 然后寻找符合条件的 j 和 k. 这样的话, 假设到 i 的节点有 m 个 (m >= 2), 那么就可以从这 m 个节点中选出 2 个来, 和 i 一起组成一个 boomerang, 那么总共有 (m * (m - 1)) 中选择.

class Solution {
private:
    int distance(pair<int, int> &p1, pair<int, int> &p2) {
        int a = p1.first - p2.first;
        int b = p1.second - p2.second;
        return a * a + b * b;
    }
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int res = 0;
        const int rows = points.size();
        for (int i = 0; i < rows; ++i) {
            // 注意此时 record 在 i 固定时产生, 第二个参数保存索引.
            unordered_map<int, vector<int>> record;
            for (int j = 0; j < rows; ++j) {
                if (j == i) continue; // 节点自身不用考虑.
                record[distance(points[j], points[i])].push_back(j);
            }
            // record 中此时保存的是到 points[i] 的距离相同的所有节点.
            for (auto &iter : record) {
                int size = iter.second.size();
                if (size >= 2) res += size * (size - 1);
            }
        }
        return res;
    }
};

C++ 实现 2

上面速度稍慢, 改成下面加快速度: 使用了关联型容器的 clear 方法.

class Solution {
private:
    int distance(pair<int, int> &p1, pair<int, int> &p2) {
        int a = p1.first - p2.first;
        int b = p1.second - p2.second;
        return a * a + b * b;
    }
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int res = 0;
        // record的第二个参数保存索引的个数
        unordered_map<int, int> record;
        const int rows = points.size();
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < rows; ++j) {
                if (j != i)
                    record[distance(points[j], points[i])] ++;
            }
            // 如果 record 中只有一个索引, 结果为 0.
            for (auto &iter : record)
                res += iter.second * (iter.second - 1);
            record.clear();
        }
        return res;
    }
};