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.
思路:
暴力求解即可通过。另外给定三点坐标,求面积公式如下:
double largestTriangleArea(vector<vector<int>>& points) { int n = points.size(); double ret = 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,area(points[i],points[j],points[k])); } } } return ret; } double area(vector<int>& a,vector<int>& b,vector<int>& c) { return 0.5*abs(a[0]*(b[1]-c[1]) +b[0]*(c[1]-a[1]) + c[0]*(a[1]-b[1]) ); }
参考: