思路： 遍历二维数组，每当遇到一个‘1’进行一次dfs，根据规则，将本次dfs到的所有元素标记为‘0’（放置重复dfs，并且能dfs到的元素一定是与当前遍历到的元素属于统一岛屿。）最后，dfs的次数就是岛屿的数量。代码如下：

class Solution {void dfs(char[][] grid, int r, int c){int nr = grid.length;int nc = grid[0].length;if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') return;grid[r][c] = '0';dfs(grid, r - 1, c);   //上下左右属于同一岛屿dfs(grid, r + 1, c);dfs(grid, r, c - 1);dfs(grid, r, c + 1);}public int numIslands(char[][] grid) {if (grid == null || grid.length == 0) return 0;int nr = grid.length;int nc = grid[0].length;int num_islands = 0;for (int r = 0; r < nr; r++){for (int c = 0; c < nc; c++){if (grid[r][c] == '1'){num_islands++;dfs(grid, r, c);}}}return num_islands;}
}