chore: update 3000. Maximum Area of Longest Diagonal Rectangle

Author: JuneLin2001

docs/LeetCode/3000. Maximum Area of Longest Diagonal Rectangle/README.mdx

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+---
+title: 3000. Maximum Area of Longest Diagonal Rectangle
+description: You are given a 2D 0-indexed integer array dimensions.
+keywords: [LeetCode, 3000. Maximum Area of Longest Diagonal Rectangle,Easy,Array]
+---
+
+# [{frontMatter.title}](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle)
+
+:::danger Easy
+:::
+
+import Tabs from "@theme/Tabs";
+import TabItem from "@theme/TabItem";
+
+<Tabs>
+  <TabItem value="description" label="題目描述" default>
+  
+## Description
+
+You are given a 2D **0-indexed** integer array dimensions.
+
+For all indices `i`, `0 <= i < dimensions.length, dimensions[i][0]` represents the length and `dimensions[i][1]` represents the width of the rectangle `i`.
+
+Return the **area** of the rectangle having the **longest** diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the **maximum** area.
+
+#### Example 1:
+
+>**Input:** dimensions = [[9,3],[8,6]]  
+**Output:** 48  
+**Explanation:**   
+For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.  
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.  
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.
+
+#### Example 2:
+
+>**Input:** dimensions = [[3,4],[4,3]]  
+**Output:** 12  
+**Explanation:** Length of diagonal is the same for both which is 5, so maximum area = 12.
+
+## Constraints:
+- `1 <= dimensions.length <= 100`
+- `dimensions[i].length == 2`
+- `1 <= dimensions[i][0], dimensions[i][1] <= 100`
+
+  </TabItem>
+  <TabItem value="solution" label="解答">
+   ## Solution
+```js
+/**
+ * @param {number[][]} dimensions
+ * @return {number}
+ */
+
+var areaOfMaxDiagonal = function (dimensions) {
+  let maxRectangleDiagonal = 0;
+  let maxRectangleArea = 0;
+
+  for (let i = 0; i < dimensions.length; i++) {
+    const length = dimensions[i][0];
+    const width = dimensions[i][1];
+    const diagonal = length * length + width * width;
+    if (diagonal > maxRectangleDiagonal) {
+      maxRectangleDiagonal = diagonal;
+      maxRectangleArea = length * width;
+    }
+    if (diagonal === maxRectangleDiagonal) {
+      maxRectangleArea = Math.max(length * width, maxRectangleArea);
+    }
+  }
+
+  return maxRectangleArea;
+};
+```
+  </TabItem>
+</Tabs>
+
+## 解題思路
+題目描述提到 `dimensions[i][0]` 會是矩形的長，然後 `dimensions[i][1]` 則是矩形的寬，需要return 對角線最長的矩形的面積。如果有多個對角線最長的矩形，則傳回面積最大的矩形的面積（ 因為 `dimensions[i].length == 2` 所以只要回傳兩個中較大的那個就好）。而矩形對角線則可以透過畢氏定理算出。
+
+以題目給的 Example 1 `dimensions = [[9,3],[8,6]]` 為例子，
+
+第一個矩形長9寬3
+ - 對角線 X：$9^2$ + $3^2$ = $X^2$ ， 所以 X = $\sqrt{90}$
+ - 面積：9 * 3 = 27
+
+第二個矩形長8寬6
+ - 對角線 X：$8^2$ + $6^2$ = $X^2$ ， 所以 X = $\sqrt{100}$
+ - 面積：8 * 6 = 48
+
+因為 $\sqrt{100}$ > $\sqrt{90}$，所以要回傳的是第一個矩形的面積 `48`。
+
+把邏輯寫成程式就是答案了！需要注意的是可能不只兩組矩形，所以需要遍歷陣列。
+
+## 心得
+順便學到了怎麼用 markdown 寫數學運算符號，好耶！
+- `$平方^2$` ： $平方^2$
+- `$\sqrt{根號}$` ： $\sqrt{根號}$

docs/LeetCode/3000. Maximum Area of Longest Diagonal Rectangle/solution.js

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+/**
+ * @param {number[][]} dimensions
+ * @return {number}
+ */
+
+var areaOfMaxDiagonal = function (dimensions) {
+  let maxRectangleDiagonal = 0;
+  let maxRectangleArea = 0;
+
+  for (let i = 0; i < dimensions.length; i++) {
+    const length = dimensions[i][0];
+    const width = dimensions[i][1];
+    const diagonal = length * length + width * width;
+    if (diagonal > maxRectangleDiagonal) {
+      maxRectangleDiagonal = diagonal;
+      maxRectangleArea = length * width;
+    }
+    if (diagonal === maxRectangleDiagonal) {
+      maxRectangleArea = Math.max(length * width, maxRectangleArea);
+    }
+  }
+
+  return maxRectangleArea;
+};