Problème
Each cell is alive (1) or dead (0) and has up to 8 neighbors. In the next generation a live cell survives with 2 or 3 live neighbors, and a dead cell becomes alive with exactly 3. Cells outside the grid are dead. All cells update at the same time.
Implement next_generation(grid).
Input
grid: A rectangular grid of 0s and 1s.
Output
Return the next generation.
Contraintes
0 ≤ rows, columns ≤ 200
Exemples
Exemple 1
Entrée :
grid = [
[0, 1, 0],
[0, 1, 0],
[0, 1, 0]
]
Sortie attendue :
expected = [
[0, 0, 0],
[1, 1, 1],
[0, 0, 0]
]Entrée :
std::vector<std::vector<long long>> grid = {
{0, 1, 0},
{0, 1, 0},
{0, 1, 0}
};
Sortie attendue :
std::vector<std::vector<long long>> expected = {
{0, 0, 0},
{1, 1, 1},
{0, 0, 0}
};Entrée :
long[][] grid = new long[][] {
new long[] {0, 1, 0},
new long[] {0, 1, 0},
new long[] {0, 1, 0}
};
Sortie attendue :
long[][] expected = new long[][] {
new long[] {0, 0, 0},
new long[] {1, 1, 1},
new long[] {0, 0, 0}
};Explication
The vertical "blinker" turns horizontal: the center survives and its left and right neighbors are born, filling the middle row.
Exemple 2
Entrée :
grid = [
[1, 1],
[1, 1]
]
Sortie attendue :
expected = [
[1, 1],
[1, 1]
]Entrée :
std::vector<std::vector<long long>> grid = {
{1, 1},
{1, 1}
};
Sortie attendue :
std::vector<std::vector<long long>> expected = {
{1, 1},
{1, 1}
};Entrée :
long[][] grid = new long[][] {
new long[] {1, 1},
new long[] {1, 1}
};
Sortie attendue :
long[][] expected = new long[][] {
new long[] {1, 1},
new long[] {1, 1}
};Explication
Every cell of the 2×2 block has 3 neighbors, so the block stays the same.
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.