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.

法1：

考虑边界位置，！+||的效果是只要有两个条件同时成立就返回，而什么情况下会两个条件都成立呢 左右两个不可能同时成立故只有左右和上下组合成立 既达成目标

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

法2：

 

道理是从重叠是的长宽边的关系来考虑

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