973. K Closest Points to Origin**
https://leetcode.com/problems/k-closest-points-to-origin/
题目描述
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 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000
C++ 实现 1
快捷的思考方法是: 保留最小的数可以使用最大堆, 而保留最大的数可以使用最小堆.
此题要求距离最小的点, 即求距离的最小值, 使用最大堆. pair
保存距离以及点的索引值, 因为最后返回的是点的坐标.
class Solution {
private:
inline int distance(const vector<int> &p) {
return p[0] * p[0] + p[1] * p[1];
}
public:
vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
priority_queue<pair<int, int>> q;
for (int i = 0; i < points.size(); ++ i) {
q.push(std::make_pair(distance(points[i]), i));
if (q.size() > K) q.pop();
}
vector<vector<int>> res;
while (!q.empty()) {
auto i = q.top().second;
res.push_back(points[i]);
q.pop();
}
return res;
}
};