[Swift]LeetCode836. 矩形重叠 | Rectangle Overlap

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10576134.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

---

矩形以列表 [x1, y1, x2, y2] 的形式表示，其中 (x1, y1) 为左下角的坐标，(x2, y2)是右上角的坐标。

如果相交的面积为正，则称两矩形重叠。需要明确的是，只在角或边接触的两个矩形不构成重叠。

给出两个矩形，判断它们是否重叠并返回结果。

示例 1：

输入：rec1 = [0,0,2,2], rec2 = [1,1,3,3]
输出：true

示例 2：

输入：rec1 = [0,0,1,1], rec2 = [1,0,2,1]
输出：false

说明：

1. 两个矩形 rec1 和 rec2 都以含有四个整数的列表的形式给出。
2. 矩形中的所有坐标都处于 -10^9 和 10^9 之间。

---

Runtime: 4 ms

Memory Usage: 19.2 MB

class Solution {
    func isRectangleOverlap(_ rec1: [Int], _ rec2: [Int]) -> Bool {
        return rec1[0] < rec2[2] && rec2[0] < rec1[2] && rec1[1] < rec2[3] && rec2[1] < rec1[3]
    }
}

---

8ms

class Solution {
    func isRectangleOverlap(_ rec1: [Int], _ rec2: [Int]) -> Bool {        
        return (min(rec1[2], rec2[2]) > max(rec1[0], rec2[0])) && 
        (min(rec1[3], rec2[3]) > max(rec1[1], rec2[1]))
    }
}