K个最近的点

Posted 2022-10-17 ssdut_yrp

给定一些 points 和一个 origin，从 points 中找到 k 个离 origin 最近的点。按照距离由小到大返回。如果两个点有相同距离，则按照x值来排序；若x值也相同，就再按照y值排序。


样例
给出 points = [[4,6],[4,7],[4,4],[2,5],[1,1]], origin = [0, 0], k = 3

返回 [[1,1],[2,5],[4,4]]

//Definition for a point.
class Point 
     int x;
     int y;
     Point()  x = 0; y = 0; 
     Point(int a, int b)  x = a; y = b; 
 
 
public class Solution 
	 /**
     * @param points a list of points
     * @param origin a point
     * @param k an integer
     * @return the k closest points
     */
    @SuppressWarnings("null")
	public static Point[] kClosest(Point[] points, Point origin, int k) 
    	Point[] outputPoints = new Point[k];
    	for(int i=0;i<points.length;i++)
    		for (int j=i;j<points.length;j++)
    			if(length(points[i],origin)>length(points[j],origin))
    				Point temp=points[i];
    				points[i]=points[j];
    				points[j]=temp;
    			else if(length(points[i],origin)==length(points[j],origin))
    				if(points[i].x>points[j].x)
    					Point temp=points[i];
        				points[i]=points[j];
        				points[j]=temp;
    				else if(points[i].y>points[j].y)
    					Point temp=points[i];
        				points[i]=points[j];
        				points[j]=temp;
    				
    			
    		
    	
    	for (int i=0;i<k;i++)
    		outputPoints[i]=points[i];
    	
		return outputPoints;
    
    
    //小到大
    public void bubbleSort(int[] unsorted)
    	for (int i=0;i<unsorted.length;i++)
    		for (int j=i;j<unsorted.length;i++)
    			if(unsorted[i]>unsorted[j])
    				int temp=unsorted[i];
    				unsorted[i]=unsorted[j];
    				unsorted[j]=temp;
    			
    
    
    
    public static double length(Point start,Point end)
    	int temp_x=Math.abs(start.x-end.x);
    	int temp_y=Math.abs(start.y-end.y);
    	return Math.sqrt(temp_x*temp_x+temp_y*temp_y);        
    
    
    public static void  print(Point pp)
    	System.out.println("最近的点：（"+pp.x+","+pp.y+")");
    

	/**
	 * @param args
	 */
	public static void main(String[] args) 
		// TODO Auto-generated method stub
		Point[] p= new Point(4,6),new Point(4,7),new Point(4,4),new Point(2,5),new Point(1,1);
		Point origin = new Point(0,0);
		int k=3;
		Point[] pNew = kClosest(p, origin, k);
		for(int i=0;i<pNew.length;i++)
			print(pNew[i]);