問題
The grid holds 1 for land and 0 for water. Land cells that touch up, down, left or right belong to the same island. Count the islands; a recursive flood fill is a natural approach.
Implement count_islands(grid).
Input
grid: A rectangular grid of 0s and 1s.
Output
Return the number of islands.
制約
0 ≤ rows, columns ≤ 100
例
例 1
入力：
grid = [
[1, 1, 0, 0],
[0, 1, 0, 1],
[0, 0, 0, 1]
]
期待される出力：
expected = 2入力：
std::vector<std::vector<long long>> grid = {
{1, 1, 0, 0},
{0, 1, 0, 1},
{0, 0, 0, 1}
};
期待される出力：
long long expected = 2;入力：
long[][] grid = new long[][] {
new long[] {1, 1, 0, 0},
new long[] {0, 1, 0, 1},
new long[] {0, 0, 0, 1}
};
期待される出力：
long expected = 2;解説
The top-left group of 3 cells is one island; the two 1s on the right form a second: 2.
例 2
入力：
grid = [
[1, 0, 1],
[0, 1, 0]
]
期待される出力：
expected = 3入力：
std::vector<std::vector<long long>> grid = {
{1, 0, 1},
{0, 1, 0}
};
期待される出力：
long long expected = 3;入力：
long[][] grid = new long[][] {
new long[] {1, 0, 1},
new long[] {0, 1, 0}
};
期待される出力：
long expected = 3;解説
Diagonal neighbors don't connect, so each 1 is its own island: 3.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。