LeetCode题解(1401)：圆和矩形是否有重叠(Python)

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题目：原题链接（中等）

标签：几何、数学

解法	时间复杂度	空间复杂度	执行用时
Ans 1 (Python)	$O (1)$	$O (1)$	44ms (37.96%)
Ans 2 (Python)
Ans 3 (Python)

解法一：

复制代码

class Solution:
    def checkOverlap(self, radius: int, x_center: int, y_center: int, x1: int, y1: int, x2: int, y2: int) -> bool:
        def distance(x, y):
            return math.sqrt(math.pow(y[1] - x[1], 2) + math.pow(y[0] - x[0], 2))

# 检查圆心是否在矩形内
        if x1 <= x_center <= x2 and y1 <= y_center <= y2:
            return True

# 检查圆的边缘是否相交
        if x_center > x2 and y1 <= y_center <= y2:
            return x_center - x2 <= radius
        if x_center < x1 and y1 <= y_center <= y2:
            return x1 - x_center <= radius
        if y_center > y2 and x1 <= x_center <= x2:
            return y_center - y2 <= radius
        if y_center < y1 and x1 <= x_center <= x2:
            return y1 - y_center <= radius

# 检查矩形的四个顶点是否在圆内
        elif (distance((x_center, y_center), (x1, y1)) <= radius or
              distance((x_center, y_center), (x1, y2)) <= radius or
              distance((x_center, y_center), (x2, y1)) <= radius or
              distance((x_center, y_center), (x2, y2)) <= radius):
            return True

return False

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class Solution:
    def checkOverlap(self, radius: int, x_center: int, y_center: int, x1: int, y1: int, x2: int, y2: int) -> bool:
        def distance(x, y):
            return math.sqrt(math.pow(y[1] - x[1], 2) + math.pow(y[0] - x[0], 2))

        # 检查圆心是否在矩形内
        if x1 <= x_center <= x2 and y1 <= y_center <= y2:
            return True

        # 检查圆的边缘是否相交
        if x_center > x2 and y1 <= y_center <= y2:
            return x_center - x2 <= radius
        if x_center < x1 and y1 <= y_center <= y2:
            return x1 - x_center <= radius
        if y_center > y2 and x1 <= x_center <= x2:
            return y_center - y2 <= radius
        if y_center < y1 and x1 <= x_center <= x2:
            return y1 - y_center <= radius

        # 检查矩形的四个顶点是否在圆内
        elif (distance((x_center, y_center), (x1, y1)) <= radius or
              distance((x_center, y_center), (x1, y2)) <= radius or
              distance((x_center, y_center), (x2, y1)) <= radius or
              distance((x_center, y_center), (x2, y2)) <= radius):
            return True

        return False