给你一个以 (radius
, x_center
, y_center
) 表示的圆和一个与坐标轴平行的矩形 (x1
, y1
, x2
, y2
),其中 (x1
, y1
) 是矩形左下角的坐标,(x2
, y2
) 是右上角的坐标。
如果圆和矩形有重叠的部分,请你返回 True ,否则返回 False 。
换句话说,请你检测是否 存在 点 (xi, yi) ,它既在圆上也在矩形上(两者都包括点落在边界上的情况)。
示例 1:
输入:radius = 1, x_center = 0, y_center = 0, x1 = 1, y1 = -1, x2 = 3, y2 = 1 输出:true 解释:圆和矩形有公共点 (1,0)
示例 2:
输入:radius = 1, x_center = 0, y_center = 0, x1 = -1, y1 = 0, x2 = 0, y2 = 1 输出:true
示例 3:
输入:radius = 1, x_center = 1, y_center = 1, x1 = -3, y1 = -3, x2 = 3, y2 = 3 输出:true
示例 4:
输入:radius = 1, x_center = 1, y_center = 1, x1 = 1, y1 = -3, x2 = 2, y2 = -1 输出:false
提示:
1 <= radius <= 2000
-10^4 <= x_center, y_center, x1, y1, x2, y2 <= 10^4
x1 < x2
y1 < y2
方法一:数学
计算矩形离圆最近的点和圆心的距离是否小于等于半径即可。
class Solution:
def checkOverlap(
self,
radius: int,
xCenter: int,
yCenter: int,
x1: int,
y1: int,
x2: int,
y2: int,
) -> bool:
dx = dy = 0
if x1 > xCenter:
dx = xCenter - x1
elif x2 < xCenter:
dx = xCenter - x2
if y1 > yCenter:
dy = yCenter - y1
elif y2 < yCenter:
dy = yCenter - y2
return dx * dx + dy * dy <= radius * radius
class Solution {
public boolean checkOverlap(
int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
int dx = x1 > xCenter ? x1 - xCenter : (x2 < xCenter ? xCenter - x2 : 0);
int dy = y1 > yCenter ? y1 - yCenter : (y2 < yCenter ? yCenter - y2 : 0);
return dx * dx + dy * dy <= radius * radius;
}
}
class Solution {
public:
bool checkOverlap(int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
int dx = x1 > xCenter ? x1 - xCenter : (x2 < xCenter ? xCenter - x2 : 0);
int dy = y1 > yCenter ? y1 - yCenter : (y2 < yCenter ? yCenter - y2 : 0);
return dx * dx + dy * dy <= radius * radius;
}
};
func checkOverlap(radius int, xCenter int, yCenter int, x1 int, y1 int, x2 int, y2 int) bool {
dx, dy := 0, 0
if x1 > xCenter {
dx = x1 - xCenter
} else if x2 < xCenter {
dx = x2 - xCenter
}
if y1 > yCenter {
dy = y1 - yCenter
} else if y2 < yCenter {
dy = y2 - yCenter
}
return dx*dx+dy*dy <= radius*radius
}