题目描述

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.)

样例

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.)

算法1

(partition select) $O(n)$

用partition找pivot，如果pivot小于k-1，右移left，否则左移right。

时间复杂度

O（n）
O（1）

参考文献

C++ 代码

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        int n=points.size(), left=0, right=n-1;
        while(left<right)
        {
            int p=partition(points, left, right);
            if(p<K-1)
                left=p+1;
            else
                right=p-1;
        }
        return vector<vector<int>>(points.begin(), points.begin()+K);
    }

    int partition(vector<vector<int>>& points, int left, int right)
    {
        int n=points.size(), p=left, l=left+1, r=right;
        while(l<=r)
        {
            if(far(points[l], points[p])&&near(points[r], points[p]))
                swap(points[l++], points[r--]);
            else if(!far(points[l], points[p]))
                l++;
            else if(!near(points[r], points[p]))
                r--;

        }
        swap(points[r], points[p]);
        return r;
    }
    bool far(vector<int> &a, vector<int> &b)
    {
        return a[0]*a[0]+a[1]*a[1]>b[0]*b[0]+b[1]*b[1];
    }
    bool near(vector<int> &a, vector<int> &b)
    {
        return a[0]*a[0]+a[1]*a[1]<b[0]*b[0]+b[1]*b[1];
    }

};