• [LC] 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.)
    
    class Solution {
        public int[][] kClosest(int[][] points, int K) {
            // keep max heap
            PriorityQueue<Cell> pq = new PriorityQueue<>((a, b) -> (b.x)*(b.x)+ (b.y)*(b.y) - (a.x)*(a.x) - (a.y)*(a.y));
    
            for (int[] point: points) {
                pq.offer(new Cell(point[0], point[1]));
                if (pq.size() > K) {
                    pq.poll();
                }
            }
            int[][] res = new int[K][2];
            for (int i = K - 1; i >= 0; i--) {
                Cell cur = pq.poll();
                res[i][0] = cur.x;
                res[i][1] = cur.y;
            }
            return res;
        }
    }
    
    class Cell {
        int x;
        int y;
        public Cell(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
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  • 原文地址:https://www.cnblogs.com/xuanlu/p/12640423.html
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