Given a circle represented as (radius, x_center, y_center) and an axis-aligned rectangle represented as (x1, y1, x2, y2), where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the rectangle.
Return True if the circle and rectangle are overlapped otherwise return False.
In other words, check if there are any point (xi, yi) such that belongs to the circle and the rectangle at the same time.
Example 1:
Input: radius = 1, x_center = 0, y_center = 0, x1 = 1, y1 = -1, x2 = 3, y2 = 1 Output: true Explanation: Circle and rectangle share the point (1,0)
Example 2:
Input: radius = 1, x_center = 0, y_center = 0, x1 = -1, y1 = 0, x2 = 0, y2 = 1 Output: true
Example 3:
Input: radius = 1, x_center = 1, y_center = 1, x1 = -3, y1 = -3, x2 = 3, y2 = 3 Output: true
Example 4:
Input: radius = 1, x_center = 1, y_center = 1, x1 = 1, y1 = -3, x2 = 2, y2 = -1 Output: false
Constraints:
1 <= radius <= 2000-10^4 <= x_center, y_center, x1, y1, x2, y2 <= 10^4x1 < x2y1 < y2
Solution: Geometry
Find the shortest distance from the center to the rectangle, return dist <= radius.
Time complexity: O(1)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: bool checkOverlap(int radius, int x_center, int y_center, int x1, int y1, int x2, int y2) { int dx = x_center - max(x1, min(x2, x_center)); int dy = y_center - max(y1, min(y2, y_center)); return dx * dx + dy * dy <= radius * radius; } }; |