Meng

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.)

Note

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

题意：在二维平面上， 给了一群点， 要求返回离原点最近的前 k 个点。 注意点之间的距离是用 Euclidean distance 来表示的。 这道题也可以用 heap 来解决。 组建一个 k 大小的 heap， 便可以在 nlogk 时间求出最结果。

代码 1

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        PriorityQueue<Pair> pq = new PriorityQueue<>(new Comparator<Pair>() {
            public int compare(Pair o1, Pair o2) {
                return o1.dist - o2.dist;
            }
        });
        for(int i = 0; i < points.length; i++) {
            pq.add(new Pair(points[i][0], points[i][1]));
        }
        int num = 0;
        int[][] ret = new int[K][2];
        while(num < K) {
            Pair cur = pq.poll();
            ret[num][0] = cur.x;
            ret[num][1] = cur.y;
            num++;
        }
        return ret;
    }
}
class Pair {
    int x;
    int y;
    int dist;
    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
        this.dist = x*x + y*y;
    }
}

代码 2 更好的解法, 参考 215 题求第k大数的解法。

class Solution {
    private Random random = new Random();
    public int[][] kClosest(int[][] points, int K) {
        if(K == points.length)
            return points;
        int start = 0, end = points.length - 1;
        int index = 0;
        while (start <= end) {
            index = partition(points, start, end);
            if (index == K) {
                break;
            }
            if (index > K) {
                end = index - 1;
            } else {
                start = index + 1;
            }
        }
        int[][] result = new int[index][2];
        for (int i = 0; i < index; i++) {
            result[i] = points[i];
        }
        return result;
    }

    private int partition(int[][] points, int start, int end) {
        int rd = start + random.nextInt(end - start + 1);
        int[] target = points[rd];
        swap(points, rd, end);
        int left = start, right = end - 1;
        while (left <= right) {
            while (left <= right && !isLarger(points[left], target)) left++;
            while (left <= right && isLarger(points[right], target)) right--;
            if (left <= right) {
                swap(points, left, right);
                left++;
                right--;
            }
        }
        swap(points, left, end);
        return left;
    }

    private void swap(int[][] points, int i1, int i2) {
        int[] temp = points[i1];
        points[i1] = points[i2];
        points[i2] = temp;
    }

    // return true if p1 dist is larger than p2
    private boolean isLarger(int[] p1, int[] p2) {
        return p1[0] * p1[0] + p1[1] * p1[1] > p2[0] * p2[0] + p2[1] * p2[1];
    }
}

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