[Solution] 973. K Closest Points to Origin

标签：derby   作者   delegate   orderby   divide   gate   其他   als   erb   

• Difficulty: Easy

Problem

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

Related Topics

Math, Divide and Conquer, Sort

Solution

解法显而易见，按照到原点的距离升序排列，取出前 K 个即可。由于 $f(x) = \sqrt{x}$ 在其定义域上单调递增，故只需要计算根号内的 $x$，即 $x^2 + y^2$ 即可。

public class Solution 
{
    public int[][] KClosest(int[][] points, int K)
    {
        Array.Sort(points, delegate (int[] x, int[] y)
        {
            return (x[0] * x[0] + x[1] * x[1]) - (y[0] * y[0] + y[1] * y[1]);
        });
        int[][] ret = new int[K][];
        Array.Copy(points, ret, K);
        return ret;
    }
}

在查看其他人的解答时，发现了如下解答（这个解答的作者的 LINQ 玩得很熟）

// no need to sqrt sinc we only want 
public class Solution 
{
    public int[][] KClosest(int[][] points, int K) 
    {
        return points
            .OrderBy(x => x[0] * x[0] + x[1] * x[1])
            .Take(K)
            .ToArray();
    }
}

[Solution] 973. K Closest Points to Origin

标签：derby   作者   delegate   orderby   divide   gate   其他   als   erb   

原文地址：https://www.cnblogs.com/Downstream-1998/p/10303109.html