Problem: 836. Rectangle Overlap 矩形重叠
解题过程
耗时100%,矩形的交集,也就是xmin、ymin的最大值,xmax、ymax的最小值,然后判断是否满足xmin < xmax且ymin < ymax即可
Code
class Solution { public: bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) { int xmi = max(rec1[0], rec2[0]); int ymi = max(rec1[1], rec2[1]); int xmx = min(rec1[2], rec2[2]); int ymx = min(rec1[3], rec2[3]); if(xmi < xmx && ymi < ymx) return true; return false; } };