LeetCode in Kotlin
812. Largest Triangle Area
Easy
Given an array of points on the X-Y plane points where points[i] = [xi, yi], return the area of the largest triangle that can be formed by any three different points. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
Output: 2.00000
Explanation: The five points are shown in the above figure. The red triangle is the largest.
Example 2:
Input: points = [[1,0],[0,0],[0,1]]
Output: 0.50000
Constraints:
3 <= points.length <= 50-50 <= xi, yi <= 50- All the given points are unique.
Solution
import kotlin.math.abs
class Solution {
fun largestTriangleArea(points: Array<IntArray>): Double {
val n = points.size
var max = 0.0
for (i in 0 until n) {
for (j in i + 1 until n) {
for (k in j + 1 until n) {
var area: Double
val a = points[i]
val b = points[j]
val c = points[k]
area = abs(area(a, b) + area(b, c) + area(c, a))
if (area > max) {
max = area
}
}
}
}
return max
}
private fun area(a: IntArray, b: IntArray): Double {
val l = b[0] - a[0]
val h = (a[1] + b[1] + 200) / 2.0
return l * h
}
}