update: 添加问题“3000.对角线最长的矩形的面积”的代码和题解 (#1099)

LetMeFly666 · web-flow · commit 76eb60f4b406 · 2025-08-26T21:48:27.000+08:00
* 3000: WA.cpp (#1098) [[6,5],[8,6],[2,10],[8,1],[9,2],[3,5],[3,5]] 20 48 * 3000: AC.cpp (#1098) - 刚刚晕了晕了 cpp - AC,100.00%,88.24% * 3000: AC.py+go+java+rust (#1098) py - AC,100.00%,82.26% go - AC,100.00%,11.11% java - AC,100.00%,33.33% rust - AC,100.00%,80.00% 中间和室友聊了一会儿哈哈 * update: 添加问题“3000.对角线最长的矩形的面积”的代码和题解 (#1099) Signed-off-by: LetMeFly666 <814114971@qq.com> --------- Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/Codes/3000-maximum-area-of-longest-diagonal-rectangle.cpp b/Codes/3000-maximum-area-of-longest-diagonal-rectangle.cpp
@@ -0,0 +1,42 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:32:11
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int areaOfMaxDiagonal(vector<vector<int>>& dimensions) {
+        int ans = 0;
+        int M = 0;
+        for (vector<int>& d : dimensions) {
+            int l2 = d[0] * d[0] + d[1] * d[1];
+            if (l2 > M) {
+                M = l2;
+                ans = d[0] * d[1];
+            } else if (l2 == M) {
+                ans = max(ans, d[0] * d[1]);
+            }
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+[[6,5],[8,6],[2,10],[8,1],[9,2],[3,5],[3,5]]
+*/
+int main() {
+    string s;
+    while (cin >> s) {
+        vector<vector<int>> v = stringToVectorVector(s);
+        Solution sol;
+        cout << sol.areaOfMaxDiagonal(v) << endl;
+    }
+    return 0;
+}
+#endif
diff --git a/Codes/3000-maximum-area-of-longest-diagonal-rectangle.go b/Codes/3000-maximum-area-of-longest-diagonal-rectangle.go
@@ -0,0 +1,21 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:36:14
+ */
+package main
+
+func areaOfMaxDiagonal(dimensions [][]int) (ans int) {
+    M := 0
+    for _, d := range dimensions {
+        l2 := d[0] * d[0] + d[1] * d[1]
+        if l2 > M {
+            M = l2
+            ans = d[0] * d[1]
+        } else if l2 == M {
+            ans = max(ans, d[0] * d[1])
+        }
+    }
+    return
+}
diff --git a/Codes/3000-maximum-area-of-longest-diagonal-rectangle.java b/Codes/3000-maximum-area-of-longest-diagonal-rectangle.java
@@ -0,0 +1,22 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:36:58
+ */
+class Solution {
+    public int areaOfMaxDiagonal(int[][] dimensions) {
+        int ans = 0;
+        int M = 0;
+        for (int[] d : dimensions) {
+            int l2 = d[0] * d[0] + d[1] * d[1];
+            if (l2 > M) {
+                M = l2;
+                ans = d[0] * d[1];
+            } else if (l2 == M) {
+                ans = Math.max(ans, d[0] * d[1]);
+            }
+        }
+        return ans;
+    }
+}
diff --git a/Codes/3000-maximum-area-of-longest-diagonal-rectangle.py b/Codes/3000-maximum-area-of-longest-diagonal-rectangle.py
@@ -0,0 +1,18 @@
+'''
+Author: LetMeFly
+Date: 2025-08-26 21:25:10
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-08-26 21:34:38
+'''
+from typing import List
+class Solution:
+    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:
+        ans = M = 0
+        for a, b in dimensions:
+            l2 = a * a + b * b
+            if l2 > M:
+                M = l2
+                ans = a * b
+            elif l2 == M:
+                ans = max(ans, a * b)
+        return ans
diff --git a/Codes/3000-maximum-area-of-longest-diagonal-rectangle.rs b/Codes/3000-maximum-area-of-longest-diagonal-rectangle.rs
@@ -0,0 +1,22 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:40:59
+ */
+impl Solution {
+    pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
+        let mut ans: i32 = 0;
+        let mut m: i32 = 0;
+        for d in dimensions.iter() {
+            let l2: i32 = d[0] * d[0] + d[1] * d[1];
+            if l2 > m {
+                m = l2;
+                ans = d[0] * d[1];
+            } else if l2 == m {
+                ans = ans.max(d[0] * d[1]);
+            }
+        }
+        ans
+    }
+}
diff --git a/Codes/lib.rs b/Codes/lib.rs
@@ -5,4 +5,4 @@
  * @LastEditTime: 2025-08-11 22:07:08
  */
 pub struct Solution;
-include!("0498-diagonal-traverse_20250825.rs");  // 这个fileName是会被脚本替换掉的
+include!("3000-maximum-area-of-longest-diagonal-rectangle.rs");  // 这个fileName是会被脚本替换掉的
diff --git a/README.md b/README.md
@@ -916,6 +916,7 @@
 |2974.最小数字游戏|简单|<a href="https://leetcode.cn/problems/minimum-number-game/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/12/LeetCode%202974.%E6%9C%80%E5%B0%8F%E6%95%B0%E5%AD%97%E6%B8%B8%E6%88%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140365205" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-number-game/solutions/2840437/letmefly-2974zui-xiao-shu-zi-you-xi-pai-ib5em/" target="_blank">LeetCode题解</a>|
 |2982.找出出现至少三次的最长特殊子字符串II|中等|<a href="https://leetcode.cn/problems/find-longest-special-substring-that-occurs-thrice-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/05/30/LeetCode%202982.%E6%89%BE%E5%87%BA%E5%87%BA%E7%8E%B0%E8%87%B3%E5%B0%91%E4%B8%89%E6%AC%A1%E7%9A%84%E6%9C%80%E9%95%BF%E7%89%B9%E6%AE%8A%E5%AD%90%E5%AD%97%E7%AC%A6%E4%B8%B2II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/139334864" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-longest-special-substring-that-occurs-thrice-ii/solutions/2795996/letmefly-2982zhao-chu-chu-xian-zhi-shao-9ubg1/" target="_blank">LeetCode题解</a>|
 |2999.统计强大整数的数目|困难|<a href="https://leetcode.cn/problems/count-the-number-of-powerful-integers/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/12/LeetCode%202999.%E7%BB%9F%E8%AE%A1%E5%BC%BA%E5%A4%A7%E6%95%B4%E6%95%B0%E7%9A%84%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147191672" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-the-number-of-powerful-integers/solutions/3649695/letmefly-2999tong-ji-qiang-da-zheng-shu-jnya8/" target="_blank">LeetCode题解</a>|
+|3000.对角线最长的矩形的面积|简单|<a href="https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/08/26/LeetCode%203000.%E5%AF%B9%E8%A7%92%E7%BA%BF%E6%9C%80%E9%95%BF%E7%9A%84%E7%9F%A9%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/150871371" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle/solutions/3764586/letmefly-3000dui-jiao-xian-zui-chang-de-cxvkc/" target="_blank">LeetCode题解</a>|
 |3011.判断一个数组是否可以变为有序|中等|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/13/LeetCode%203011.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%E6%98%AF%E5%90%A6%E5%8F%AF%E4%BB%A5%E5%8F%98%E4%B8%BA%E6%9C%89%E5%BA%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140391465" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/solutions/2841665/letmefly-3011pan-duan-yi-ge-shu-zu-shi-f-i9ck/" target="_blank">LeetCode题解</a>|
 |3019.按键变更的次数|简单|<a href="https://leetcode.cn/problems/number-of-changing-keys/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/07/LeetCode%203019.%E6%8C%89%E9%94%AE%E5%8F%98%E6%9B%B4%E7%9A%84%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144983704" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-changing-keys/solutions/3040949/letmefly-3019an-jian-bian-geng-de-ci-shu-pgzx/" target="_blank">LeetCode题解</a>|
 |3024.三角形类型|简单|<a href="https://leetcode.cn/problems/type-of-triangle/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/19/LeetCode%203024.%E4%B8%89%E8%A7%92%E5%BD%A2%E7%B1%BB%E5%9E%8B/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/148061722" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/type-of-triangle/solutions/3680826/letmefly-3024san-jiao-xing-lei-xing-shou-q46h/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 3000.对角线最长的矩形的面积.md b/Solutions/LeetCode 3000.对角线最长的矩形的面积.md
@@ -0,0 +1,213 @@
+---
+title: 3000.对角线最长的矩形的面积：一次遍历
+date: 2025-08-26 21:41:54
+tags: [题解, LeetCode, 简单, 数组, 遍历, 模拟, 数学]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】3000.对角线最长的矩形的面积：一次遍历
+
+力扣题目链接：[https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle/](https://leetcode.cn/problems/maximum-area-of-longest-diagonal-rectangle/)
+
+<p>给你一个下标从<strong> 0</strong> 开始的二维整数数组 <code>dimensions</code>。</p>
+
+<p>对于所有下标 <code>i</code>（<code>0 &lt;= i &lt; dimensions.length</code>），<code>dimensions[i][0]</code> 表示矩形 <span style="font-size: 13.3333px;"> <code>i</code></span> 的长度，而 <code>dimensions[i][1]</code> 表示矩形 <span style="font-size: 13.3333px;"> <code>i</code></span> 的宽度。</p>
+
+<p>返回对角线最 <strong>长 </strong>的矩形的<strong> 面积 </strong>。如果存在多个对角线长度相同的矩形，返回面积最<strong> 大 </strong>的矩形的面积。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>dimensions = [[9,3],[8,6]]
+<strong>输出：</strong>48
+<strong>解释：</strong>
+下标 = 0，长度 = 9，宽度 = 3。对角线长度 = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈<!-- notionvc: 882cf44c-3b17-428e-9c65-9940810216f1 --> 9.487。
+下标 = 1，长度 = 8，宽度 = 6。对角线长度 = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10。
+因此，下标为 1 的矩形对角线更长，所以返回面积 = 8 * 6 = 48。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>dimensions = [[3,4],[4,3]]
+<strong>输出：</strong>12
+<strong>解释：</strong>两个矩形的对角线长度相同，为 5，所以最大面积 = 12。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= dimensions.length &lt;= 100</code></li>
+	<li><code>dimensions[i].length == 2</code></li>
+	<li><code>1 &lt;= dimensions[i][0], dimensions[i][1] &lt;= 100</code></li>
+</ul>
+
+
+    
+## 解题方法：遍历维护最大值
+
+使用一个变量l2维护最大的对角线长度的平方（避免无意义的开根号），遍历所有矩形：
+
++ 如果当前矩形对角线长度的平方大于l2，则更新l2并直接更新答案为当前矩形的面积；
++ 否则如果当前对角线长度的平方等于l2，则更新答案为当前矩形面积和答案的最大值。
+
+### 时空复杂度分析
+
++ 时间复杂度$O(len(dimensions))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:32:11
+ */
+class Solution {
+public:
+    int areaOfMaxDiagonal(vector<vector<int>>& dimensions) {
+        int ans = 0;
+        int M = 0;
+        for (vector<int>& d : dimensions) {
+            int l2 = d[0] * d[0] + d[1] * d[1];
+            if (l2 > M) {
+                M = l2;
+                ans = d[0] * d[1];
+            } else if (l2 == M) {
+                ans = max(ans, d[0] * d[1]);
+            }
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+[[6,5],[8,6],[2,10],[8,1],[9,2],[3,5],[3,5]]
+*/
+int main() {
+    string s;
+    while (cin >> s) {
+        vector<vector<int>> v = stringToVectorVector(s);
+        Solution sol;
+        cout << sol.areaOfMaxDiagonal(v) << endl;
+    }
+    return 0;
+}
+#endif
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-08-26 21:25:10
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-08-26 21:34:38
+'''
+from typing import List
+class Solution:
+    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:
+        ans = M = 0
+        for a, b in dimensions:
+            l2 = a * a + b * b
+            if l2 > M:
+                M = l2
+                ans = a * b
+            elif l2 == M:
+                ans = max(ans, a * b)
+        return ans
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:36:58
+ */
+class Solution {
+    public int areaOfMaxDiagonal(int[][] dimensions) {
+        int ans = 0;
+        int M = 0;
+        for (int[] d : dimensions) {
+            int l2 = d[0] * d[0] + d[1] * d[1];
+            if (l2 > M) {
+                M = l2;
+                ans = d[0] * d[1];
+            } else if (l2 == M) {
+                ans = Math.max(ans, d[0] * d[1]);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:36:14
+ */
+package main
+
+func areaOfMaxDiagonal(dimensions [][]int) (ans int) {
+    M := 0
+    for _, d := range dimensions {
+        l2 := d[0] * d[0] + d[1] * d[1]
+        if l2 > M {
+            M = l2
+            ans = d[0] * d[1]
+        } else if l2 == M {
+            ans = max(ans, d[0] * d[1])
+        }
+    }
+    return
+}
+```
+
+#### Rust
+
+```rust
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-08-26 21:25:10
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-08-26 21:40:59
+ */
+impl Solution {
+    pub fn area_of_max_diagonal(dimensions: Vec<Vec<i32>>) -> i32 {
+        let mut ans: i32 = 0;
+        let mut m: i32 = 0;
+        for d in dimensions.iter() {
+            let l2: i32 = d[0] * d[0] + d[1] * d[1];
+            if l2 > m {
+                m = l2;
+                ans = d[0] * d[1];
+            } else if l2 == m {
+                ans = ans.max(d[0] * d[1]);
+            }
+        }
+        ans
+    }
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/150871371)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/08/26/LeetCode%203000.%E5%AF%B9%E8%A7%92%E7%BA%BF%E6%9C%80%E9%95%BF%E7%9A%84%E7%9F%A9%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
