自定义排序函数

最新推荐文章于 2023-10-11 22:52:30 发布

独孤金泽

最新推荐文章于 2023-10-11 22:52:30 发布

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本文链接：https://blog.youkuaiyun.com/qq_35747066/article/details/90049844

本文介绍了一种算法，用于在二维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几里得距离，并使用比较函数进行排序，最终返回前K个最近的点坐标。

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973. K Closest Points to Origin

Medium

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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:

class Solution {
public:
    static bool Cmp(const pair<vector<int>,int>& a,const pair<vector<int>,int>& b){
        return a.second < b.second;
    }
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        vector<pair<vector<int>,int>> Result;
        for(auto point : points){
            int distance = point[0] * point[0] + point[1] * point[1];
            Result.push_back(make_pair(point,distance));
        }
        sort(Result.begin(),Result.end(),Cmp);
        vector<vector<int>> Res;
        for(int i = 0;i < K;i ++){
            vector<int> tmp = Result[i].first;
            Res.push_back(tmp);
        }
        return Res;
    }
    
};

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.)