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:

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

 

Approach #1: Math. [Java]

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return rec1[0] < rec2[2] && rec2[0] < rec1[2] && rec1[1] < rec2[3] && rec2[1] < rec1[3];
    }
}

　　

Analysis:

Before we do it in 2D plane, let's try it in 1D.

Given 2 segment (left1, right1), (left2, right2), how can we check whether they overlap?

If these two intervals overlaps, it should exits an number x,

left1 < x < right1 && left2 < x < right2

left1 < x < right2 && left2 < x < right1

left1 < right2 && left2 < right1

This is the sufficient and necessary condition for two segment overlap.

Explanation:

For 2D, if two rectancle overlap both on x and y,  they overlap in the plane.

 

Analysis:

https://leetcode.com/problems/rectangle-overlap/discuss/132340/C%2B%2BJavaPython-1-line-Solution-1D-to-2D