线性扫描二维网格图,如果一个节点包含'1',那么它是触发深度优先搜索的根节点。在 DFS 期间,每个访问过的节点都应设置为“0”以标记为访问过的节点。计算触发 DFS 的根节点的数量,这个数字就是岛的数量,因为从某个根开始的每个 DFS 标识一个岛。
using System; namespace ConsoleApplication{ public class Matrix{ public int PrintNumberOfIslands(char[,] grid){ bool[,] visited = new bool[grid.GetLength(0), grid.GetLength(1)]; int res = 0; for (int i = 0; i < grid.GetLength(0); i++){ for (int j = 0; j < grid.GetLength(1); j++){ if (grid[i, j] == '1' && !visited[i, j]){ DFS(grid, visited, i, j); res++; } } } return res; } public void DFS(char[,] grid, bool[,] visited, int i, int j){ if (i < 0 || i >= grid.GetLength(0)) return; if (j < 0 || j >= grid.GetLength(1)) return; if (grid[i, j] != '1' || visited[i, j]) return; visited[i, j] = true; DFS(grid, visited, i + 1, j); DFS(grid, visited, i - 1, j); DFS(grid, visited, i, j + 1); DFS(grid, visited, i, j - 1); } } class Program{ static void Main(string[] args){ Matrix m = new Matrix(); char[,] mm = { { '1', '1', '1', '1', '0' }, { '1', '1', '0', '1', '0' }, { '1', '1', '0', '0', '0' }, { '0', '0', '0', '0', '1' } }; Console.WriteLine(m.PrintNumberOfIslands(mm)); } } }输出结果
2