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910db9a | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 | # 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)
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