973. K Closest Points to Origin

[Tcdian]

973. K Closest Points to Origin

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。

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.)

Example 3

输入：N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
输出：false

Note

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

[Tcdian]

Solution

使用 JavaScript 自带的排序，时间复杂度 O(NlogN)，其中 N 是 points 的长度
还可以使用堆，时间复杂度 O(NlogK)。(太长了，懒得写...)

• JavaScript Solution

/**
 * @param {number[][]} points
 * @param {number} K
 * @return {number[][]}
 */
var kClosest = function(points, K) {
    points.sort(([x1, y1], [x2, y2]) => x1 * x1 + y1 * y1 - (x2 * x2 + y2 * y2));
    return points.slice(0, K);
};

• TypeScript Solution

var kClosest = function(points: number[][], K: number): number[][] {
    points.sort(([x1, y1], [x2, y2]) => x1 * x1 + y1 * y1 - (x2 * x2 + y2 * y2));
    return points.slice(0, K);
};