2023-06-25每日一题

一、题目编号

1401. 圆和矩形是否有重叠

二、题目链接

点击跳转到题目位置

三、题目描述

给你一个以 (radius, xCenter, yCenter) 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，而 (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）。

提示：

1 <= radius <= 2000
-10⁴ <= xCenter, yCenter <= 10⁴
-10⁴ <= x1 < x2 <= 10⁴
-10⁴ <= y1 < y2 <= 10⁴

四、解题代码

class Solution {long long distance(int xCenter, int yCenter, int x0, int y0){return (long long) pow(xCenter - x0, 2) + (long long) pow(yCenter - y0, 2);}public:bool checkOverlap(int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {if (x1 <= xCenter && xCenter <= x2 && y1 <= yCenter && yCenter <= y2) {return true;}if (x1 <= xCenter && xCenter <= x2 && y2 <= yCenter && yCenter <= y2 + radius) {return true;}if (x1 <= xCenter && xCenter <= x2 && y1 - radius <= yCenter && yCenter <= y1) {return true;}if (x1 - radius <= xCenter && xCenter <= x1 && y1 <= yCenter && yCenter <= y2) {return true;}if (x2 <= xCenter && xCenter <= x2 + radius && y1 <= yCenter && yCenter <= y2) {return true;}if(distance(xCenter, yCenter, x1, y2) <= radius * radius){//左上角return true;}if(distance(xCenter, yCenter, x2, y2) <= radius * radius){//右上角return true;}if(distance(xCenter, yCenter, x2, y1) <= radius * radius){//右下角return true;}if(distance(xCenter, yCenter, x1, y1) <= radius * radius){//左下角return true;}return false;}
};