【leetcode】【easy】【每日一题】836. Rectangle Overlap

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本文探讨了如何判断两个轴对齐的矩形是否重叠的问题，通过将区间问题转化为坐标轴上的投影，提供了一种简洁高效的解决方案。文章详细解析了判断逻辑，并附带LeetCode题目链接与代码实现。

836. Rectangle Overlap

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.

题目链接：https://leetcode-cn.com/problems/rectangle-overlap/

思路

本来做题思路是列举所有true的情况，后来发现事情么有那么简单，于是看到了一个分享：

https://leetcode-cn.com/problems/rectangle-overlap/solution/tu-jie-jiang-ju-xing-zhong-die-wen-ti-zhuan-hua-we/

这个思路是将区间问题转到了在坐标轴上的投影，感觉很巧妙。

class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        if(rec1.size()<4 || rec2.size()<4) return false;
        if (rec1[2]<=rec2[0] || rec1[0]>=rec2[2]||rec1[1]>=rec2[3]|| rec1[3]<=rec2[1]) return false;
        else return true;
    }
};