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  • LeetCode 973. K Closest Points to Origin

    原题链接在这里:https://leetcode.com/problems/k-closest-points-to-origin/

    题目:

    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. 1 <= K <= points.length <= 10000
    2. -10000 < points[i][0] < 10000
    3. -10000 < points[i][1] < 10000

    题解:

    Could use maxHeap to track maximum distance, keep adding, when size > K, poll.

    Time Complexity: O(nlogK). n = points.length.

    Space: O(K).

    AC Java:

     1 class Solution {
     2     public int[][] kClosest(int[][] points, int K) {
     3         if(points == null || points.length < K){
     4             return points;
     5         }
     6         
     7         if(K <= 0){
     8             return new int[0][2];
     9         }
    10         
    11         PriorityQueue<int []> maxHeap = new PriorityQueue<>((a, b) -> b[2] - a[2]);
    12         for(int [] p : points){
    13             int dis = p[0] * p[0] + p[1] * p[1];
    14             maxHeap.add(new int[]{p[0], p[1], dis});
    15             if(maxHeap.size() > K){
    16                 maxHeap.poll();
    17             }
    18         }
    19         
    20         int [][] res = new int[maxHeap.size()][2];
    21         for(int i = 0; i < res.length; i++){
    22             int [] cur = maxHeap.poll();
    23             res[i] = new int[]{cur[0], cur[1]};
    24         }
    25         
    26         return res;
    27     }
    28 }

    类似Top K Frequent Elements.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12216495.html
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