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
A:
就是用heap ,, 选择了max heap, 这样,可以只用 K个。 然后,每次加入时候,对比头部
class Solution { public: vector<vector<int>> kClosest(vector<vector<int>>& points, int K) { priority_queue<vector<int>, vector<vector<int>>, comparator> maxHeap; for(auto & p : points){ if(maxHeap.size()==K ){ auto & t = maxHeap.top(); if(t[0]*t[0] + t[1]*t[1] > p[0]*p[0] + p[1]*p[1]){ maxHeap.pop(); maxHeap.push(p); } }else{ maxHeap.push(p); } } vector<vector<int>> res; while(not maxHeap.empty()){ auto t = maxHeap.top(); maxHeap.pop(); res.push_back(t); } return res; } private: struct comparator{ bool operator()(const vector<int> & a, const vector<int> &b){ return a[0]*a[0] + a[1]*a[1] < b[0]*b[0] + b[1]*b[1]; } }; };
Mistakes:
auto & t = maxHeap.top();
maxheap.pop();
一开始,我写成这样,想重复 利用这个空间来着。 然而 t 可能已经被销毁了。会有heap error
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