# 【LeetCode】 973. K Closest Points to Origin

## Description
> 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.)

> Note:
> * `1 <= K <= points.length <= 10000`
> * `-10000 < points[i][0] < 10000`
> * `-10000 < points[i][1] < 10000`

> 在平面上我們擁有一個列表的點。找到前`K`個最靠近原點`(0, 0)`的點。
> (在這裡，兩點的距離定義為歐氏距離。)
> 你用任何順序可以回傳答案。答案保證是唯一的(除了不同排序)。

> 提示：
> * `1 <= K <= points.length <= 10000`
> * `-10000 < points[i][0] < 10000`
> * `-10000 < points[i][1] < 10000`

## Example:

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.)

## Solution

* 將所有點排序，比較function自己重寫，定義為與原點的距離。
* 原點的距離是$\sqrt{x^2 + y^2}$，但為了節省計算時間可以不用開根號。
* 排序完畢之後，使用`resize()`拿出前`K`項。

### Code

bool myCompare (vector<int> p1,vector<int> p2) {
        return (p1[0] * p1[0] + p1[1] * p1[1] <
                p2[0] * p2[0] + p2[1] * p2[1]); 
    }

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        sort(points.begin(), points.end(), myCompare);
        points.resize(K);
        return points;
    }
};

###### tags: `LeetCode` `C++`