LeetCode 973. K Closest Points to Origin

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.)

这道题目是一个很经典的问题，即找到前k个最小的元素。这里的大小由该点到原点的距离来定义。

-------------------------------------------------------------------------------------------------------------------------------------

首先第一种办法：最简单的，上来就排序。排完序之后取前k个就行了

代码如下：

其中用到了java 8中的->来实现comparator的传递

    public int[][] kClosest(int[][] points, int K) {
        Arrays.sort( points,(p1,p2)->p1[0]*p1[0]+p1[1]*p1[1]-p2[0]*p2[0]-p2[1]*p2[1] );
        return Arrays.copyOfRange(points,0,K);
    }

---------------------------------------------------------------------------------------------------------------------------------

第二种办法，调用优先队列，维持一个容量为k的最大堆（堆顶元素为最大值）

对于进来的每一个元素，先将该元素插入最大堆（时间复杂度为O（lgK））,然后将最大堆顶元素扔掉（常数时间）。

即每个操作为O(lgK) 

总共N个元素，故为O（NlgK）

代码如下

    public int[][] kClosest(int[][] points, int K) {
        //优先队列默认是最小堆，这里通过comparator设为最大堆（堆顶为最大值）
        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[] point:points){
            pq.offer(point);//将元素插入优先队列
            if(pq.size()>K){
                pq.poll();//扔掉队列里的最大元素
            }
        }
        int[][] ret = new int[K][2];
        while(K>0){
            ret[--K] = pq.poll();
        }
        return ret;
    }

-------------------------------------------------------------------------------------------------------------------------------

第三种方法：基于切分的方法。

类似于快速排序中的切分

import java.util.*;

class Solution { 
    
    public int[][] kClosest(int[][] points, int K) {
        int low = 0;
        int high = points.length-1;
        while(low<high){
            int j = partition(points,low,high);
            if(j==K){
                break;
            }
            else if(j<K){
                low = j+1;
            }
            else if(j>K){
                high = j-1;
            }
        }
        return Arrays.copyOfRange(points, 0, K);
    }

    private int partition(int[][] points,int low,int high) {
        int i = low;
        int j= high+1;
        while(true){
            while( less(points[++i],points[low]) ){
                if(i==high) break;
            }
            while( less(points[low],points[--j]) ){
                // if(j==low) break;
            }
            if(i>=j){
                break;
            }
            swap(points,i,j);
        }
        swap(points,low,j);
        return j;
        
    }
    
    private boolean less(int[] p1, int[] p2) {
        return (p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1])<0;
    }
    
    
    private void swap(int[][] points,int i ,int j){
        int[] t = points[i];
        points[i] = points[j];
        points[j] = t;
    }
    
    private void shuffle(int[][] points) {
        Random random = new Random();
        for(int ind = 1; ind < points.length; ind++) {
            int r = random.nextInt(ind + 1);
            swap(points, ind, r);
        }
    }
}