题目：

Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).

The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2).

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]]
Explanation: The answer [[-2,4],[3,3]] would also be accepted.

题解：

//参考kanishkmewal
class Solution {
    public int[][] kClosest(int[][] points, int k) {
        int lo = 0;
        int hi = points.length - 1;
        int pivotIdx = -1;
        // 寻找前k个最近的
        while (pivotIdx != k - 1) {
            // 返回本次排序好的位置
            pivotIdx = partition(points, lo, hi);
            // 检查位置与K值的关系
            if (pivotIdx < k) {
                lo = pivotIdx + 1;
            } else {
                hi = pivotIdx - 1;
            }
        }
        return Arrays.copyOfRange(points, 0, k);
    }
    
    private int partition(int[][] points, int lo, int hi) {
        double pivotDist = dist(points[hi]);
        int j = lo;
        for (int i = lo, k = hi; j < k; ) {
            double dist = dist(points[j]);
            if (dist < pivotDist) {
                swap(points, i++, j++);
            } else if (dist > pivotDist) {
                swap(points, --k, j);
            } else {
                j++;
            }
        }
        swap(points, j, hi);
        return j;
    }
    
    private double dist(int[] p) {
        return p[0] * p[0] + p[1] * p[1];
    }
    
    private void swap(int[][] points, int i, int j) {
        int[] tmp = points[i];
        points[i] = points[j];
        points[j] = tmp;
    }
}

思路：

quickSelect