Problème
The factory method pattern hides how objects are created: you ask the factory for a shape by name and get back an object that knows how to compute its own area.
Each description is a shape name followed by its numbers, separated by spaces:
circle r: area π·r²square s: area s²rectangle w h: area w·htriangle b h: area b·h / 2
Implement shape_factory_areas(shapes).
Input
shapes: The descriptions. Numbers may have decimals.
Output
Return the area of each shape, in order. Answers within 1e-6 are accepted.
Contraintes
1 ≤ number of shapes ≤ 1,000
0 < every number ≤ 1,000
Exemples
Entrée :
shapes = ["square 3", "rectangle 2 5", "triangle 4 3"]
Sortie attendue :
expected = [9.0, 10.0, 6.0]
Entrée :
std::vector<std::string> shapes = {"square 3", "rectangle 2 5", "triangle 4 3"};
Sortie attendue :
std::vector<double> expected = {9.0, 10.0, 6.0};
Entrée :
string[] shapes = new string[] {"square 3", "rectangle 2 5", "triangle 4 3"};
Sortie attendue :
double[] expected = new double[] {9.0, 10.0, 6.0};
Explication
3² = 9, 2·5 = 10 and 4·3/2 = 6.
Entrée :
shapes = ["circle 1"]
Sortie attendue :
expected = [3.141592653589793]
Entrée :
std::vector<std::string> shapes = {"circle 1"};
Sortie attendue :
std::vector<double> expected = {3.141592653589793};
Entrée :
string[] shapes = new string[] {"circle 1"};
Sortie attendue :
double[] expected = new double[] {3.141592653589793};
Explication
π·1² = π ≈ 3.141593.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
Read the first word to know which shape to build.
Indice 2
Map each shape name to its area formula.
Indice 3
Use π from your language's math library and keep full precision.
Complexité attendue
O(number of shapes) time, O(number of shapes) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.