73 lines
2.6 KiB
C++
73 lines
2.6 KiB
C++
// Source : https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/
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// Author : Hao Chen
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// Date : 2021-04-20
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/*****************************************************************************************************
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*
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* You are given an array points where points[i] = [xi, yi] is the coordinates of the i^th point on a
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* 2D plane. Multiple points can have the same coordinates.
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*
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* You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at
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* (xj, yj) with a radius of rj.
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*
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* For each query queries[j], compute the number of points inside the j^th circle. Points on the
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* border of the circle are considered inside.
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*
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* Return an array answer, where answer[j] is the answer to the j^th query.
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*
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* Example 1:
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*
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* Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]]
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* Output: [3,2,2]
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* Explanation: The points and circles are shown above.
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* queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.
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*
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* Example 2:
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*
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* Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]]
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* Output: [2,3,2,4]
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* Explanation: The points and circles are shown above.
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* queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.
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*
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* Constraints:
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*
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* 1 <= points.length <= 500
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* points[i].length == 2
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* 0 <= xi, yi <= 500
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* 1 <= queries.length <= 500
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* queries[j].length == 3
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* 0 <= xj, yj <= 500
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* 1 <= rj <= 500
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* All coordinates are integers.
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*
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* Follow up: Could you find the answer for each query in better complexity than O(n)?
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******************************************************************************************************/
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class Solution {
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private:
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//refer to: https://stackoverflow.com/a/7227057/13139221
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bool inCircle( vector<int>& point, vector<int>& circle ) {
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int x = point[0], y = point[1];
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int xo = circle[0], yo = circle[1], r = circle[2];
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int dx = abs(x-xo);
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if ( dx > r ) return false;
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int dy = abs(y-yo);
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if ( dy > r ) return false;
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if ( dx + dy <= r ) return true;
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return ( dx*dx + dy*dy <= r*r );
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}
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public:
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vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
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vector<int> result;
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for(auto& c : queries) {
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int cnt = 0;
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for(auto& p : points) {
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if ( inCircle(p, c) ) cnt++;
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}
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result.push_back(cnt);
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}
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return result;
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}
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};
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