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| # 963. Minimum Area Rectangle II | ||
| You are given an array of points in the **X-Y** plane `points` where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>. | ||
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| Return *the minimum area of any rectangle formed from these points, with sides **not necessarily parallel** to the X and Y axes*. If there is not any such rectangle, return `0`. | ||
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| Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted. | ||
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| #### Example 1: | ||
|  | ||
| <pre> | ||
| <strong>Input:</strong> points = [[1,2],[2,1],[1,0],[0,1]] | ||
| <strong>Output:</strong> 2.00000 | ||
| <strong>Explanation:</strong> The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2. | ||
| </pre> | ||
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| #### Example 2: | ||
|  | ||
| <pre> | ||
| <strong>Input:</strong> points = [[0,1],[2,1],[1,1],[1,0],[2,0]] | ||
| <strong>Output:</strong> 1.00000 | ||
| <strong>Explanation:</strong> The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1. | ||
| </pre> | ||
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| #### Example 3: | ||
|  | ||
| <pre> | ||
| <strong>Input:</strong> points = [[0,3],[1,2],[3,1],[1,3],[2,1]] | ||
| <strong>Output:</strong> 0 | ||
| <strong>Explanation:</strong> There is no possible rectangle to form from these points. | ||
| </pre> | ||
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| #### Constraints: | ||
| * `1 <= points.length <= 50` | ||
| * `points[i].length == 2` | ||
| * <code>0 <= x<sub>i</sub>, y<sub>i</sub> <= 4 * 10<sup>4</sup></code> | ||
| * All the given points are **unique**. | ||
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| ## Solutions (Rust) | ||
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| ### 1. Solution | ||
| ```Rust | ||
| use std::collections::HashSet; | ||
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| impl Solution { | ||
| pub fn min_area_free_rect(points: Vec<Vec<i32>>) -> f64 { | ||
| let points_set = points.iter().map(|p| (p[0], p[1])).collect::<HashSet<_>>(); | ||
| let mut ret = f64::NAN; | ||
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| for i in 0..points.len() { | ||
| let (xi, yi) = (points[i][0], points[i][1]); | ||
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| for j in i + 1..points.len() { | ||
| let (xj, yj) = (points[j][0], points[j][1]); | ||
| let ij2 = (xi - xj).pow(2) as f64 + (yi - yj).pow(2) as f64; | ||
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| for k in j + 1..points.len() { | ||
| let (xk, yk) = (points[k][0], points[k][1]); | ||
| let ik2 = (xi - xk).pow(2) as f64 + (yi - yk).pow(2) as f64; | ||
| let jk2 = (xj - xk).pow(2) as f64 + (yj - yk).pow(2) as f64; | ||
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| if ij2 + ik2 == jk2 && points_set.contains(&(xj + xk - xi, yj + yk - yi)) { | ||
| ret = ret.min(ij2.sqrt() * ik2.sqrt()); | ||
| } else if ij2 + jk2 == ik2 && points_set.contains(&(xi + xk - xj, yi + yk - yj)) | ||
| { | ||
| ret = ret.min(ij2.sqrt() * jk2.sqrt()); | ||
| } else if ik2 + jk2 == ij2 && points_set.contains(&(xi + xj - xk, yi + yj - yk)) | ||
| { | ||
| ret = ret.min(ik2.sqrt() * jk2.sqrt()); | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| if ret.is_nan() { | ||
| return 0.; | ||
| } | ||
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| ret | ||
| } | ||
| } | ||
| ``` |
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| # 963. 最小面积矩形 II | ||
| 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。 | ||
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| 如果没有任何矩形，就返回 0。 | ||
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| #### 示例 1: | ||
|  | ||
| <pre> | ||
| <strong>输入:</strong> points = [[1,2],[2,1],[1,0],[0,1]] | ||
| <strong>输出:</strong> 2.00000 | ||
| <strong>解释:</strong> 最小面积的矩形出现在 [1,2],[2,1],[1,0],[0,1] 处，面积为 2。 | ||
| </pre> | ||
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| #### 示例 2: | ||
|  | ||
| <pre> | ||
| <strong>输入:</strong> points = [[0,1],[2,1],[1,1],[1,0],[2,0]] | ||
| <strong>输出:</strong> 1.00000 | ||
| <strong>解释:</strong> 最小面积的矩形出现在 [1,0],[1,1],[2,1],[2,0] 处，面积为 1。 | ||
| </pre> | ||
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| #### 示例 3: | ||
|  | ||
| <pre> | ||
| <strong>输入:</strong> points = [[0,3],[1,2],[3,1],[1,3],[2,1]] | ||
| <strong>输出:</strong> 0 | ||
| <strong>解释:</strong> 没法从这些点中组成任何矩形。 | ||
| </pre> | ||
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| #### 示例 4: | ||
|  | ||
| <pre> | ||
| <strong>输入:</strong> [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]] | ||
| <strong>输出:</strong> 2.00000 | ||
| <strong>解释:</strong> 最小面积的矩形出现在 [2,1],[2,3],[3,3],[3,1] 处，面积为 2。 | ||
| </pre> | ||
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| #### 提示: | ||
| 1. `1 <= points.length <= 50` | ||
| 2. `0 <= points[i][0] <= 40000` | ||
| 3. `0 <= points[i][1] <= 40000` | ||
| 4. 所有的点都是不同的。 | ||
| 5. 与真实值误差不超过 `10^-5` 的答案将视为正确结果。 | ||
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| ## 题解 (Rust) | ||
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| ### 1. 题解 | ||
| ```Rust | ||
| use std::collections::HashSet; | ||
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| impl Solution { | ||
| pub fn min_area_free_rect(points: Vec<Vec<i32>>) -> f64 { | ||
| let points_set = points.iter().map(|p| (p[0], p[1])).collect::<HashSet<_>>(); | ||
| let mut ret = f64::NAN; | ||
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| for i in 0..points.len() { | ||
| let (xi, yi) = (points[i][0], points[i][1]); | ||
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| for j in i + 1..points.len() { | ||
| let (xj, yj) = (points[j][0], points[j][1]); | ||
| let ij2 = (xi - xj).pow(2) as f64 + (yi - yj).pow(2) as f64; | ||
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| for k in j + 1..points.len() { | ||
| let (xk, yk) = (points[k][0], points[k][1]); | ||
| let ik2 = (xi - xk).pow(2) as f64 + (yi - yk).pow(2) as f64; | ||
| let jk2 = (xj - xk).pow(2) as f64 + (yj - yk).pow(2) as f64; | ||
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| if ij2 + ik2 == jk2 && points_set.contains(&(xj + xk - xi, yj + yk - yi)) { | ||
| ret = ret.min(ij2.sqrt() * ik2.sqrt()); | ||
| } else if ij2 + jk2 == ik2 && points_set.contains(&(xi + xk - xj, yi + yk - yj)) | ||
| { | ||
| ret = ret.min(ij2.sqrt() * jk2.sqrt()); | ||
| } else if ik2 + jk2 == ij2 && points_set.contains(&(xi + xj - xk, yi + yj - yk)) | ||
| { | ||
| ret = ret.min(ik2.sqrt() * jk2.sqrt()); | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| if ret.is_nan() { | ||
| return 0.; | ||
| } | ||
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| ret | ||
| } | ||
| } | ||
| ``` |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,39 @@ | ||
| use std::collections::HashSet; | ||
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| impl Solution { | ||
| pub fn min_area_free_rect(points: Vec<Vec<i32>>) -> f64 { | ||
| let points_set = points.iter().map(|p| (p[0], p[1])).collect::<HashSet<_>>(); | ||
| let mut ret = f64::NAN; | ||
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| for i in 0..points.len() { | ||
| let (xi, yi) = (points[i][0], points[i][1]); | ||
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| for j in i + 1..points.len() { | ||
| let (xj, yj) = (points[j][0], points[j][1]); | ||
| let ij2 = (xi - xj).pow(2) as f64 + (yi - yj).pow(2) as f64; | ||
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| for k in j + 1..points.len() { | ||
| let (xk, yk) = (points[k][0], points[k][1]); | ||
| let ik2 = (xi - xk).pow(2) as f64 + (yi - yk).pow(2) as f64; | ||
| let jk2 = (xj - xk).pow(2) as f64 + (yj - yk).pow(2) as f64; | ||
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| if ij2 + ik2 == jk2 && points_set.contains(&(xj + xk - xi, yj + yk - yi)) { | ||
| ret = ret.min(ij2.sqrt() * ik2.sqrt()); | ||
| } else if ij2 + jk2 == ik2 && points_set.contains(&(xi + xk - xj, yi + yk - yj)) | ||
| { | ||
| ret = ret.min(ij2.sqrt() * jk2.sqrt()); | ||
| } else if ik2 + jk2 == ij2 && points_set.contains(&(xi + xj - xk, yi + yj - yk)) | ||
| { | ||
| ret = ret.min(ik2.sqrt() * jk2.sqrt()); | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| if ret.is_nan() { | ||
| return 0.; | ||
| } | ||
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| ret | ||
| } | ||
| } |
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