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

## 973. K Closest Points to Origin

## 题目地址

<https://leetcode.com/problems/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.)
```

## 代码

### Approach 1: Sort

```java
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: max-Heap

```java
public int[][] kClosest(int[][] points, int K) {
    PriorityQueue<int[]> pq = new PriorityQueue<int[]>((p1, p2) -> p2[0] * p2[0] + p2[1] * p2[1] - p1[0] * p1[0] - p1[1] * p1[1]);
    for (int[] p : points) {
        pq.offer(p);
        if (pq.size() > K) {
            pq.poll();
        }
    }
    int[][] res = new int[K][2];
    while (K > 0) {
        res[--K] = pq.poll();
    }
    return res;
}
```

### Approach 3: Quick sort

the average time complexity is **O(N)** , but just like quick sort, in the worst case, this solution would be degenerated to **O(N^2)**, and **pratically**, the real time it takes on leetcode is **15ms**.

```java
public int[][] kClosest(int[][] points, int K) {
    int len =  points.length, l = 0, r = len - 1;
    while (l <= r) {
        int mid = helper(points, l, r);
        if (mid == K) break;
        if (mid < K) {
            l = mid + 1;
        } else {
            r = mid - 1;
        }
    }
    return Arrays.copyOfRange(points, 0, K);
}

private int helper(int[][] A, int l, int r) {
    int[] pivot = A[l];
    while (l < r) {
        while (l < r && compare(A[r], pivot) >= 0) r--;
        A[l] = A[r];
        while (l < r && compare(A[l], pivot) <= 0) l++;
        A[r] = A[l];
    }
    A[l] = pivot;
    return l;
}

private int compare(int[] p1, int[] p2) {
    return p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1];
}
```


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