973. K Closest Points to Origin-优快云博客

本文介绍了一种算法，用于在平面上找到距离原点最近的K个点。通过计算每个点到原点的欧氏距离，并使用映射表按距离排序，从而有效地找到最接近原点的指定数量的点。

We have a list of points on the plane. Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation: 
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000

Approach: native. [C++]

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        map<double, pair<int, int>> temp;
        for (int i = 0; i < points.size(); ++i) {
            double dis = sqrt(points[i][0] * points[i][0] + points[i][1] * points[i][1]);
            temp[dis] = {points[i][0], points[i][1]};
        }
        vector<vector<int>> ans;
        map<double, pair<int, int>>::iterator it;
        it = temp.begin();
        while (K--)
            ans.push_back({it->second.first, it->second.second}), it++;
        return ans;
    }
};

转载于:https://www.cnblogs.com/ruruozhenhao/p/10344390.html