You are given a circle represented as (radius, xCenter, yCenter) 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 is any point (xi, yi) that belongs to the circle and the rectangle at the same time.
Example 1:
Input: radius = 1, xCenter = 0, yCenter = 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, xCenter = 1, yCenter = 1, x1 = 1, y1 = -3, x2 = 2, y2 = -1 Output: false
Example 3:
Input: radius = 1, xCenter = 0, yCenter = 0, x1 = -1, y1 = 0, x2 = 0, y2 = 1 Output: true
Constraints:
1 <= radius <= 2000-104 <= xCenter, yCenter <= 104-104 <= x1 < x2 <= 104-104 <= y1 < y2 <= 104class Solution {
public boolean checkOverlap(int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
// Circle inside Rectangle
if (x1 <= xCenter && x2 >= xCenter && y1 <= yCenter && y2 >= yCenter)
return true;
// Circle on a side of Rectangle
if (x1 <= xCenter && x2 >= xCenter && y2 <= yCenter && yCenter - y2 <= radius) // top
return true;
if (x1 <= xCenter && x2 >= xCenter && y1 >= yCenter && y1 - yCenter <= radius) // bottom
return true;
if (y1 <= yCenter && y2 >= yCenter && x1 >= xCenter && x1 - xCenter <= radius) // left
return true;
if (y1 <= yCenter && y2 >= yCenter && x2 <= xCenter && xCenter - x2 <= radius) // left
return true;
// Circle in corner of Rectangle cases
if (distSquare(xCenter, yCenter, x1, y1) <= radius * radius) // bottom left
return true;
if (distSquare(xCenter, yCenter, x2, y2) <= radius * radius) // top right
return true;
if (distSquare(xCenter, yCenter, x1, y2) <= radius * radius) // top left
return true;
if (distSquare(xCenter, yCenter, x2, y1) <= radius * radius) // bottom right
return true;
return false;
}
long distSquare(int x1, int y1, int x2, int y2) {
return (long) Math.pow(x1 - x2, 2) + (long) Math.pow(y1 - y2, 2);
}
}