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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.

• Java
• C++
• Python
• Go

• 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;
      }
  }
  

• // 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;
      }
  };
  

• 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
  

• 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
  }
  

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