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Formatted question description: https://leetcode.ca/all/812.html
812. Largest Triangle Area (Easy)
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.
Companies:
Google
Related Topics:
Math
Solution 1.
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class Solution { public double largestTriangleArea(int[][] points) { double largestArea = 0; int length = points.length; for (int i = 0; i < length; i++) { int[] point1 = points[i]; for (int j = i + 1; j < length; j++) { int[] point2 = points[j]; for (int k = j + 1; k < length; k++) { int[] point3 = points[k]; if (sameLine(point1, point2, point3)) continue; double area = getArea(point1, point2, point3); largestArea = Math.max(largestArea, area); } } } return largestArea; } public boolean sameLine(int[] point1, int[] point2, int[] point3) { int delta1X = point2[0] - point1[0], delta1Y = point2[1] - point1[1]; int delta2X = point3[0] - point2[0], delta2Y = point3[1] - point2[1]; return delta1X * delta2Y == delta2X * delta1Y; } public double getArea(int[] point1, int[] point2, int[] point3) { double side1 = getDistance(point1, point2); double side2 = getDistance(point2, point3); double side3 = getDistance(point3, point1); double halfPerimeter = (side1 + side2 + side3) / 2; double area = Math.sqrt(halfPerimeter * (halfPerimeter - side1) * (halfPerimeter - side2) * (halfPerimeter - side3)); return area; } public double getDistance(int[] point1, int[] point2) { return Math.sqrt(((double) point2[0] - (double) point1[0]) * ((double) point2[0] - (double) point1[0]) + ((double) point2[1] - (double) point1[1]) * ((double) point2[1] - (double) point1[1])); } } ############ class Solution { public double largestTriangleArea(int[][] points) { double ans = 0; for (int[] p1 : points) { int x1 = p1[0], y1 = p1[1]; for (int[] p2 : points) { int x2 = p2[0], y2 = p2[1]; for (int[] p3 : points) { int x3 = p3[0], y3 = p3[1]; int u1 = x2 - x1, v1 = y2 - y1; int u2 = x3 - x1, v2 = y3 - y1; double t = Math.abs(u1 * v2 - u2 * v1) / 2.0; ans = Math.max(ans, t); } } } return ans; } }
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// OJ: https://leetcode.com/problems/largest-triangle-area/ // Time: O(N^3) // Space: O(1) // Ref: https://leetcode.com/problems/largest-triangle-area/solution/ class Solution { private: double area (vector<int> &a, vector<int> &b, vector<int> &c) { return .5 * abs(a[0]*b[1] + b[0]*c[1] + c[0]*a[1] -a[1]*b[0] - b[1]*c[0] - c[1]*a[0]); } public: double largestTriangleArea(vector<vector<int>>& points) { double ans = 0; for (int i = 0, N = points.size(); i < N; ++i) { for (int j = i + 1; j < N; ++j) { for (int k = j + 1; k < N; ++k) { ans = max(ans, area(points[i], points[j], points[k])); } } } return ans; } };
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class Solution: def largestTriangleArea(self, points: List[List[int]]) -> float: ans = 0 for x1, y1 in points: for x2, y2 in points: for x3, y3 in points: u1, v1 = x2 - x1, y2 - y1 u2, v2 = x3 - x1, y3 - y1 t = abs(u1 * v2 - u2 * v1) / 2 ans = max(ans, t) return ans ############ class Solution: def largestTriangleArea(self, points): """ :type points: List[List[int]] :rtype: float """ res = 0 N = len(points) for i in range(N - 2): for j in range(i + 1, N - 1): for k in range(i + 2, N): (x1, y1), (x2, y2), (x3, y3) = points[i], points[j], points[k] res = max(res, 0.5 * abs(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2))) return res
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func largestTriangleArea(points [][]int) float64 { ans := 0.0 for _, p1 := range points { x1, y1 := p1[0], p1[1] for _, p2 := range points { x2, y2 := p2[0], p2[1] for _, p3 := range points { x3, y3 := p3[0], p3[1] u1, v1 := x2-x1, y2-y1 u2, v2 := x3-x1, y3-y1 t := float64(abs(u1*v2-u2*v1)) / 2.0 ans = math.Max(ans, t) } } } return ans } func abs(x int) int { if x < 0 { return -x } return x }