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# 973.K-Closest-Points-to-Origin

## 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.)
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

## 代码

### Approach #1 Sort

Time: O(NlogN) && Space: O(N)
class Solution {
public int[][] kClosest(int[][] points, int K) {
int N = points.length;
int[] dists = new int[N];
for (int i = 0; i < N; ++i)
dists[i] = dist(points[i]);
Arrays.sort(dists);
int distK = dists[K-1];
int[][] ans = new int[K][2];
int t = 0;
for (int i = 0; i < N; ++i)
if (dist(points[i]) <= distK)
ans[t++] = points[i];
return ans;
}
public int dist(int[] point) {
return point[0] * point[0] + point[1] * point[1];
}
}

### Approach #2 Divide and Conquer

Time: O(NlogN) && Space: O(N)
class Solution {
int[][] points;
public int[][] kClosest(int[][] points, int K) {
this.points = points;
int n = points.length;
divideConquer(0, n-1, K);
return Arrays.copyOfRange(points, 0, K);
}
public void divideConquer(int start, int end, int K) {
if (start >= end) return;
int mid = partion(start, end);
if (mid < K) {
divideConquer(mid + 1, end, K);
} else if (mid > K) {
divideConquer(start, mid - 1, K);
}
}
public int partion(int lo, int hi) {
int pivot = dist(hi);
int s = lo;
for (int i = lo; i < hi; i++) {
if (dist(i) < pivot) {
swap(i, s);
s++;
}
}
swap(hi, s);
return s;
}
private void swap(int i, int j) {
int[] temp = points[i];
points[i] = points[j];
points[j] = temp;
}
public int dist(int i) {
return points[i][0] * points[i][0] + points[i][1] * points[i][1];
}
}