libgdx判断actor与circle是否重叠

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libgdx判断actor与circle是否重叠

实质是检测矩形与circle是否重叠基本函数,判断点是否在circle中 public static boolean IsInside( float x, float y, Circle circle ){ float disX = x - circle.x; float disY = y - circle.y; return disX*disX + disY*disY <= circle.radius*circle.radius; } 再判断矩形4个点是否在其中,有一个在即重叠 public…

libgdx学习记录25——Rectangle与Circle是否重叠

Rect与Circle重叠有三种情况: 1. Rect至少有一个角在Circle里面 2. Circle与Rect的左边或右边相交,或者Circle在Rect内 3. Circle与Rect的顶边或底边相交,或者Circle在Rect内判断代码: public static boolean IsInside( float x, float y, Circle circle ){ float disX = x - circle.x; float disY = y - circle.y; retu…

libgdx判断矩形重叠碰撞

有两种方式. 1. 排除法,排除四种不可能重叠的情况就是了. public static boolean IsOverlap( Rectangle rect1, Rectangle rect2 ){ float x1 = rect1.x, y1 = rect1.y, w1 = rect1.width, h1 = rect1.height; float x2 = rect2.x, y2 = rect2.y, w2 = rect2.width, h2 = rect2.height; return !…

Java判断多个时间段是否重叠(重叠区间个数)

import java.util.ArrayList; import java.util.Collections; import java.util.List; /** * 判断多个时间段是否出现重叠 * @author cavancao */ public class TimeSlotUtil { public static boolean checkOverlap(List<String> list){ Collections.sort(list);//排序ASC boolean flag…

Overlapping rectangles判断两个矩形是否重叠的问题 C++

Given two rectangles, find if the given two rectangles overlap or not. A rectangle is denoted by providing the x and y co-ordinates of two points: the left top corner and the right bottom corner of the rectangle. Note that two rectangles sharing a si…

判断两个控件在同一个Window上是否有重叠

判断两个控件在同一个Window上是否有重叠 //对UIView写分类 - (BOOL)intersectWithView:(UIView *)view; - (BOOL)intersectWithView:(UIView *)view { UIWindow *window = [UIApplication sharedApplication].keyWindow; CGRect selfRect = [self convertRect:self.bounds toView:window]; C…