给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。
岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外,你可以假设该网格的四条边均被水包围。
示例 1:
输入:grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ] 输出:1
示例 2:
输入:grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ] 输出:3
public class Solution { private int n, m; public int numIslands(char[][] grid) { n = grid.length; if (n == 0) return 0; m = grid[0].length; int num = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) if (grid[i][j] == '1') { DFS(grid, i, j); ++num; } } return num; } private void DFS(char[][] grid, int i, int j) { if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != '1') return; grid[i][j] = '0'; DFS(grid, i + 1, j); DFS(grid, i - 1, j); DFS(grid, i, j + 1); DFS(grid, i, j - 1); } }