第2天了。
今天的题目是[https://leetcode.com/problems/k-closest-points-to-origin/](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:
1Input: points = [[1,3],[-2,2]], K = 12Output: [[-2,2]]3Explanation:4The distance between (1, 3) and the origin is sqrt(10).5The distance between (-2, 2) and the origin is sqrt(8).6Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.7We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].
Example 2:
1Input: points = [[3,3],[5,-1],[-2,4]], K = 22Output: [[3,3],[-2,4]]3(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
今天的题目比较简单,虽然是一道Mediem
的题目,但是不知道为什么好像常规做法就AC了。解法就是算每个点到原点的距离先,然后排序,最后取出前K个就好了:
1vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {2 vector<vector<int>> res;3 vector<pair<int, int>> index;4 for(int i = 0, size = points.size(); i < size; i++) {5 pair<int,int> p = {i, points[i][0]*points[i][0] + points[i][1]*points[i][1]};6 index.push_back(p);7 }8
9 sort(index.begin(), index.end(), [](const pair<int, int> &pi, const pair<int, int> &pj) {10 return pi.second < pj.second;11 });12
13 for(int i = 0;i < K;i++) {14 res.push_back(points[index[i].first]);15 }3 collapsed lines
16
17 return res;18}