上篇文章与大家一起看了题目介绍并留了独立思考的时间给大家。解决方案方法枚举思路和算法关于求解三角形面积的公式可以参考百度百科「三角形面积公式」。我们可以枚举所有的三角形然后计算三角形的面积并找出最大的三角形面积。设三角形三个顶点的坐标为 (x1 ,y1)、(x2 ,y2) 和 (x3 ,y3)则三角形面积 S 可以用行列式的绝对值表示代码Python3class Solution: def largestTriangleArea(self, points: List[List[int]]) - float: def triangleArea(x1: int, y1: int, x2: int, y2: int, x3: int, y3: int) - float: return abs(x1 * y2 x2 * y3 x3 * y1 - x1 * y3 - x2 * y1 - x3 * y2) / 2 return max(triangleArea(x1, y1, x2, y2, x3, y3) for (x1, y1), (x2, y2), (x3, y3) in combinations(points, 3))Cclass Solution { public: double triangleArea(int x1, int y1, int x2, int y2, int x3, int y3) { return 0.5 * abs(x1 * y2 x2 * y3 x3 * y1 - x1 * y3 - x2 * y1 - x3 * y2); } double largestTriangleArea(vectorvectorint points) { int n points.size(); double ret 0.0; for (int i 0; i n; i) { for (int j i 1; j n; j) { for (int k j 1; k n; k) { ret max(ret, triangleArea(points[i][0], points[i][1], points[j][0], points[j][1], points[k][0], points[k][1])); } } } return ret; } };Javaclass Solution { public double largestTriangleArea(int[][] points) { int n points.length; double ret 0.0; for (int i 0; i n; i) { for (int j i 1; j n; j) { for (int k j 1; k n; k) { ret Math.max(ret, triangleArea(points[i][0], points[i][1], points[j][0], points[j][1], points[k][0], points[k][1])); } } } return ret; } public double triangleArea(int x1, int y1, int x2, int y2, int x3, int y3) { return 0.5 * Math.abs(x1 * y2 x2 * y3 x3 * y1 - x1 * y3 - x2 * y1 - x3 * y2); } }
