| # You have a list of points in the plane. Return the area of the largest | |
| # triangle that can be formed by any 3 of the points. | |
| # Example: | |
| # Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]] | |
| # Output: 2 | |
| # Explanation: The five points are show in the figure below. The red triangle | |
| # is the largest. | |
| # Notes: | |
| # 3 <= points.length <= 50. | |
| # No points will be duplicated. | |
| # -50 <= points[i][j] <= 50. | |
| # Answers within 10^-6 of the true value will be accepted as correct. | |
| class Solution: | |
| def largestTriangleArea(self, points): | |
| """ | |
| :type points: List[List[int]] | |
| :rtype: float | |
| """ | |
| from math import sqrt | |
| def dist(p1, p2): | |
| return sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) | |
| def area(d1, d2, d3): | |
| h = (d1 + d2 + d3) / 2.0 | |
| return sqrt(abs(h * (h - d1) * (h - d2) * (h - d3))) | |
| max_area = 0 | |
| n = len(points) | |
| for i in range(n): | |
| p1 = points[i] | |
| for j in range(i + 1, n): | |
| p2 = points[j] | |
| for k in range(j + 1, n): | |
| p3 = points[k] | |
| d1 = dist(p1, p2) | |
| d2 = dist(p2, p3) | |
| d3 = dist(p3, p1) | |
| a = area(d1, d2, d3) | |
| if a > max_area: | |
| max_area = a | |
| return max_area | |
| sol = Solution().largestTriangleArea | |
| print(sol([[0, 0], [0, 1], [1, 0], [0, 2], [2, 0]]), 2) | |