Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.
If there isn‘t any rectangle, return 0.
Example 1:
Input: [[1,2],[2,1],[1,0],[0,1]] Output: 2.00000 Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.
Example 2:
Input: [[0,1],[2,1],[1,1],[1,0],[2,0]] Output: 1.00000 Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.
Example 3:
Input: [[0,3],[1,2],[3,1],[1,3],[2,1]] Output: 0 Explanation: There is no possible rectangle to form from these points.
Example 4:
Input: [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]] Output: 2.00000 Explanation: The minimum area rectangle occurs at [2,1],[2,3],[3,3],[3,1], with an area of 2.
Note:
1 <= points.length <= 500 <= points[i][0] <= 400000 <= points[i][1] <= 40000- All points are distinct.
- Answers within
10^-5of the actual value will be accepted as correct.
给定在 xy 平面上的一组点，确定由这些点组成的任何矩形的最小面积，其中矩形的边不一定平行于 x 轴和 y 轴。
如果没有任何矩形，就返回 0。
示例 1：
输入：[[1,2],[2,1],[1,0],[0,1]] 输出：2.00000 解释：最小面积的矩形出现在 [1,2],[2,1],[1,0],[0,1] 处，面积为 2。
示例 2：
输入：[[0,1],[2,1],[1,1],[1,0],[2,0]] 输出：1.00000 解释：最小面积的矩形出现在 [1,0],[1,1],[2,1],[2,0] 处，面积为 1。
示例 3：
输入：[[0,3],[1,2],[3,1],[1,3],[2,1]] 输出：0 解释：没法从这些点中组成任何矩形。
示例 4：
输入：[[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]] 输出：2.00000 解释：最小面积的矩形出现在 [2,1],[2,3],[3,3],[3,1] 处，面积为 2。
提示：
1 <= points.length <= 500 <= points[i][0] <= 400000 <= points[i][1] <= 40000- 所有的点都是不同的。
- 与真实值误差不超过
10^-5的答案将视为正确结果。
460ms
1 class Solution {
2 func minAreaFreeRect(_ points: [[Int]]) -> Double {
3 var ans:Double = -1
4 var set:Set<Int> = Set<Int>()
5 for p in points
6 {
7 set.insert((p[0] << 16) | p[1])
8 }
9
10 var n:Int = points.count
11 for i in 0..<n
12 {
13 for j in 0..<n
14 {
15 var x0:Int = points[j][0] - points[i][0]
16 var y0:Int = points[j][1] - points[i][1]
17 if i != j
18 {
19 for k in 0..<n
20 {
21 if i != k && j != k
22 {
23 var x1:Int = points[k][0] - points[i][0]
24 var y1:Int = points[k][1] - points[i][1]
25 if x0 * x1 + y0 * y1 == 0
26 {
27 var x:Int = points[j][0] + points[k][0] - points[i][0]
28 var y:Int = points[j][1] + points[k][1] - points[i][1]
29 if set.contains((x << 16) | y)
30 {
31 var cur:Double = sqrt(Double(x0 * x0 + y0 * y0)) * sqrt(Double(x1 * x1 + y1 * y1))
32 if ans < 0 || ans > cur
33 {
34 ans = cur
35 }
36 }
37 }
38 }
39 }
40 }
41 }
42 }
43 return ans < 0 ? 0 : ans
44 }
45 }
原文地址：https://www.cnblogs.com/strengthen/p/10165340.html
时间： 2024-10-08 16:06:01