Code & Fun

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:

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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:

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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

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:

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vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
vector<vector<int>> res;
vector<pair<int, int>> index;
for(int i = 0, size = points.size(); i < size; i++) {
pair<int,int> p = {i, points[i][0]*points[i][0] + points[i][1]*points[i][1]};
index.push_back(p);
}

sort(index.begin(), index.end(), [](const pair<int, int> &pi, const pair<int, int> &pj) {
return pi.second < pj.second;
});

for(int i = 0;i < K;i++) {
res.push_back(points[index[i].first]);
}

return res;
}

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