|
| 1 | +--- |
| 2 | +title: 812.最大三角形面积：三角形面积公式考察（附三种公式） |
| 3 | +date: 2025-09-27 15:19:16 |
| 4 | +tags: [题解, LeetCode, 简单, 几何, 数组, 数学] |
| 5 | +categories: [题解, LeetCode] |
| 6 | +--- |
| 7 | + |
| 8 | +# 【LetMeFly】812.最大三角形面积：三角形面积公式考察（附三种公式） |
| 9 | + |
| 10 | +力扣题目链接：[https://leetcode.cn/problems/largest-triangle-area/](https://leetcode.cn/problems/largest-triangle-area/) |
| 11 | + |
| 12 | +<p>给你一个由 <strong>X-Y</strong> 平面上的点组成的数组 <code>points</code> ，其中 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> 。从其中取任意三个不同的点组成三角形，返回能组成的最大三角形的面积。与真实值误差在 <code>10<sup>-5</sup></code> 内的答案将会视为正确答案<strong>。</strong></p> |
| 13 | + |
| 14 | +<p> </p> |
| 15 | + |
| 16 | +<p><strong class="example">示例 1：</strong></p> |
| 17 | +<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png" style="height: 369px; width: 450px;" /> |
| 18 | +<pre> |
| 19 | +<strong>输入：</strong>points = [[0,0],[0,1],[1,0],[0,2],[2,0]] |
| 20 | +<strong>输出：</strong>2.00000 |
| 21 | +<strong>解释：</strong>输入中的 5 个点如上图所示，红色的三角形面积最大。 |
| 22 | +</pre> |
| 23 | + |
| 24 | +<p><strong class="example">示例 2：</strong></p> |
| 25 | + |
| 26 | +<pre> |
| 27 | +<strong>输入：</strong>points = [[1,0],[0,0],[0,1]] |
| 28 | +<strong>输出：</strong>0.50000 |
| 29 | +</pre> |
| 30 | + |
| 31 | +<p> </p> |
| 32 | + |
| 33 | +<p><strong>提示：</strong></p> |
| 34 | + |
| 35 | +<ul> |
| 36 | + <li><code>3 <= points.length <= 50</code></li> |
| 37 | + <li><code>-50 <= x<sub>i</sub>, y<sub>i</sub> <= 50</code></li> |
| 38 | + <li>给出的所有点 <strong>互不相同</strong></li> |
| 39 | +</ul> |
| 40 | + |
| 41 | + |
| 42 | + |
| 43 | +## 解题方法：暴力 |
| 44 | + |
| 45 | +三角形面积公式怎么算，有这些方法： |
| 46 | + |
| 47 | +假设二维平面三角形三个顶点的坐标 $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$， |
| 48 | + |
| 49 | +1. 行列式公式（Shoelace公式）：$S = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$ |
| 50 | + |
| 51 | +2. 向量叉积公式：设 $\vec{AB} = (x_2-x_1, y_2-y_1), \vec{AC} = (x_3-x_1, y_3-y_1)$，则 |
| 52 | + $S = \frac{1}{2} | \vec{AB} \times \vec{AC} | = \frac{1}{2} | (x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1) |$ |
| 53 | + |
| 54 | +3. 海伦公式（已知三边长 $a,b,c$）：$a = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2},\ b = \sqrt{(x_1-x_3)^2 + (y_1-y_3)^2},\ c = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$ |
| 55 | + $s = \frac{a+b+c}{2},\ S = \sqrt{s(s-a)(s-b)(s-c)}$ |
| 56 | + |
| 57 | + |
| 58 | ++ 时间复杂度$O(len(points)^3)$ |
| 59 | ++ 空间复杂度$O(1)$ |
| 60 | + |
| 61 | +### AC代码 |
| 62 | + |
| 63 | +#### C++ |
| 64 | + |
| 65 | +```cpp |
| 66 | +/* |
| 67 | + * @Author: LetMeFly |
| 68 | + * @Date: 2025-09-27 12:53:08 |
| 69 | + * @LastEditors: LetMeFly.xyz |
| 70 | + * @LastEditTime: 2025-09-27 15:10:05 |
| 71 | + */ |
| 72 | +class Solution { |
| 73 | +private: |
| 74 | + inline int calc(vector<vector<int>>& points, int i, int j, int k) { |
| 75 | + int x1 = points[i][0], y1 = points[i][1]; |
| 76 | + int x2 = points[j][0], y2 = points[j][1]; |
| 77 | + int x3 = points[k][0], y3 = points[k][1]; |
| 78 | + return abs((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)); |
| 79 | + } |
| 80 | +public: |
| 81 | + double largestTriangleArea(vector<vector<int>>& points) { |
| 82 | + int ans = 0; |
| 83 | + for (int i = 0; i < points.size(); i++) { |
| 84 | + for (int j = i + 1; j < points.size(); j++) { |
| 85 | + for (int k = j + 1; k < points.size(); k++) { |
| 86 | + ans = max(ans, calc(points, i, j, k)); |
| 87 | + } |
| 88 | + } |
| 89 | + } |
| 90 | + return double(ans) / 2; |
| 91 | + } |
| 92 | +}; |
| 93 | +``` |
| 94 | +
|
| 95 | +#### Python |
| 96 | +
|
| 97 | +```python |
| 98 | +''' |
| 99 | +Author: LetMeFly |
| 100 | +Date: 2025-09-27 12:53:08 |
| 101 | +LastEditors: LetMeFly.xyz |
| 102 | +LastEditTime: 2025-09-27 15:15:55 |
| 103 | +''' |
| 104 | +from typing import List |
| 105 | +from itertools import combinations |
| 106 | +
|
| 107 | +class Solution: |
| 108 | + def largestTriangleArea(self, points: List[List[int]]) -> float: |
| 109 | + ans = 0 |
| 110 | + for p1, p2, p3 in combinations(points, 3): |
| 111 | + x1, y1 = p2[0] - p1[0], p2[1] - p1[1] |
| 112 | + x2, y2 = p3[0] - p1[0], p3[1] - p1[1] |
| 113 | + ans = max(ans, abs(x1 * y2 - y1 * x2)) # 记得abs |
| 114 | + return ans / 2 |
| 115 | +``` |
| 116 | + |
| 117 | +#### Python一行版 |
| 118 | + |
| 119 | +```python |
| 120 | +''' |
| 121 | +Author: LetMeFly |
| 122 | +Date: 2025-09-27 12:53:08 |
| 123 | +LastEditors: LetMeFly.xyz |
| 124 | +LastEditTime: 2025-09-27 15:17:57 |
| 125 | +''' |
| 126 | +from typing import List |
| 127 | +from itertools import combinations |
| 128 | + |
| 129 | +class Solution: |
| 130 | + def largestTriangleArea(self, points: List[List[int]]) -> float: |
| 131 | + return max(abs((p2[0] - p1[0]) * (p3[1] - p1[1]) - (p2[1] - p1[1]) * (p3[0] - p1[0])) for p1, p2, p3 in combinations(points, 3)) / 2 |
| 132 | +``` |
| 133 | + |
| 134 | +> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/152167745)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/2025/09/27/LeetCode%200812.%E6%9C%80%E5%A4%A7%E4%B8%89%E8%A7%92%E5%BD%A2%E9%9D%A2%E7%A7%AF/)哦~ |
| 135 | +> |
| 136 | +> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode) |
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