[Swift]LeetCode836. 矩形重疊 | Rectangle Overlap

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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.

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

若是相交的面積爲正，則稱兩矩形重疊。須要明確的是，只在角或邊接觸的兩個矩形不構成重疊。

給出兩個矩形，判斷它們是否重疊並返回結果。

示例 1：

輸入：rec1 = [0,0,2,2], rec2 = [1,1,3,3]
輸出：true

示例 2：

輸入：rec1 = [0,0,1,1], rec2 = [1,0,2,1]
輸出：false

說明：

1. 兩個矩形 rec1 和 rec2 都以含有四個整數的列表的形式給出。
2. 矩形中的全部座標都處於 -10^9 和 10^9 之間。

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Runtime: 4 ms

Memory Usage: 19.2 MB

1 class Solution {
2     func isRectangleOverlap(_ rec1: [Int], _ rec2: [Int]) -> Bool {
3         return rec1[0] < rec2[2] && rec2[0] < rec1[2] && rec1[1] < rec2[3] && rec2[1] < rec1[3]
4     }
5 }

---

8ms

1 class Solution {
2     func isRectangleOverlap(_ rec1: [Int], _ rec2: [Int]) -> Bool {        
3         return (min(rec1[2], rec2[2]) > max(rec1[0], rec2[0])) && 
4         (min(rec1[3], rec2[3]) > max(rec1[1], rec2[1]))
5     }
6 }