LeetCode #836 - Rectangle Overlap

最新推荐文章于 2024-11-18 11:49:25 发布

LawFile

最新推荐文章于 2024-11-18 11:49:25 发布

阅读量122

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/88386959

LeetCode 专栏收录该内容

625 篇文章

订阅专栏

本文介绍了一种用于判断两个轴对齐矩形是否重叠的高效算法。通过比较两个矩形的边界坐标，该算法可以快速确定它们是否有交集。适用于计算机图形学、游戏开发及地图应用等领域。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

A rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) are the coordinates of its bottom-left corner, and (x2, y2) are the coordinates of its top-right corner.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two (axis-aligned) rectangles, return whether they overlap.

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

Notes:

Both rectangles rec1 and rec2 are lists of 4 integers.
All coordinates in rectangles will be between -10^9 and 10^9.

class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        return max(rec1[0],rec2[0])<min(rec1[2],rec2[2])&&max(rec1[1],rec2[1])<min(rec1[3],rec2[3]);
    }
};