K Closest Points to Origin 最接近原点的 K 个点（Medium）（JAVA）

【LeetCode】 973. K Closest Points to Origin 最接近原点的 K 个点（Medium）（JAVA）

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.)

Note:

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

题目大意

我们有一个由平面上的点组成的列表 points。需要从中找出 K 个距离原点 (0, 0) 最近的点。

（这里，平面上两点之间的距离是欧几里德距离。）

你可以按任何顺序返回答案。除了点坐标的顺序之外，答案确保是唯一的。

解题方法

一般计算 K 个最值问题都用堆排序，下面用两种排序算法说明下为什么堆排序更快

插入排序

1. 采用插入排序的方法
2. 时间复杂度 O(nlogK)

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        List<int[]> list = new ArrayList<>();
        for (int i = 0; i < points.length; i++) {
            insert(list, points, i, K);
        }
        int[][] res = new int[K][2];
        for (int i = 0; i < list.size(); i++) {
            res[i] = points[list.get(i)[0]];
        }
        return res;
    }

    // list[0]: index, list[1]: distance
    public void insert(List<int[]> list, int[][] points, int index, int K) {
        int sum = points[index][0] * points[index][0] + points[index][1] * points[index][1];
        if (list.size() >= K && sum > list.get(list.size() - 1)[1]) return;
        if (list.size() >= K) list.remove(list.size() - 1);
        int start = 0;
        int end = list.size() - 1;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (list.get(mid)[1] == sum) {
                start = mid;
                break;
            } else if (list.get(mid)[1] > sum) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        list.add(start, new int[]{index, sum});
    }
}

执行耗时:52 ms,击败了14.21% 的Java用户
内存消耗:47.3 MB,击败了40.28% 的Java用户

堆排序

1. 采用堆排序的方法
2. 时间复杂度 O(Klogn)：因为 K < n，所以 Klogn < nlogK，堆排序更快

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        PriorityQueue<int[]> queue = new PriorityQueue<>((a, b) -> (b[1] - a[1]));
        for (int i = 0; i < K; i++) {
            queue.offer(new int[]{i, points[i][0] * points[i][0] + points[i][1] * points[i][1]});
        }
        for (int i = K; i < points.length; i++) {
            int dis = points[i][0] * points[i][0] + points[i][1] * points[i][1];
            if (dis < queue.peek()[1]) {
                queue.poll();
                queue.offer(new int[]{i, dis});
            }
        }
        int[][] res = new int[K][2];
        for (int i = 0; i < K; i++) {
            res[i] = points[queue.poll()[0]];
        }
        return res;
    }
}

执行耗时:31 ms,击败了56.28% 的Java用户
内存消耗:47.6 MB,击败了21.72% 的Java用户

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本文地址：https://blog.csdn.net/qq_16927853/article/details/109569591