问题：
给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请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，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
提示：
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'
解答思路：
以下是使用深度优先搜索（DFS）解决岛屿数量问题的 Java 代码：
class Solution {public int numIslands(char[][] grid) {if (grid == null || grid.length == 0) {return 0;}int numIslands = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == '1') {numIslands++;dfs(grid, i, j);}}}return numIslands;}private void dfs(char[][] grid, int i, int j) {if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') {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);}public static void main(String[] args) {char[][] grid1 = {{'1', '1', '1', '1', '0'},{'1', '1', '0', '1', '0'},{'1', '1', '0', '0', '0'},{'0', '0', '0', '0', '0'}};char[][] grid2 = {{'1', '1', '0', '0', '0'},{'1', '1', '0', '0', '0'},{'0', '0', '1', '0', '0'},{'0', '0', '0', '1', '1'}};Solution solution = new Solution();System.out.println("示例 1 中岛屿的数量: " + solution.numIslands(grid1));System.out.println("示例 2 中岛屿的数量: " + solution.numIslands(grid2));}
}这段代码的思路是：遍历整个二维网格，如果遇到值为 '1' 的单元格，就将其标记为已访问（将 '1' 改为 '0'），然后对其周围的单元格进行深度优先搜索，将与之相连的陆地都标记为已访问。这样，每次遇到一个未访问过的 '1' 单元格，就表示找到了一个新的岛屿，岛屿数量加 1。
(文章为作者在学习java过程中的一些个人体会总结和借鉴，如有不当、错误的地方，请各位大佬批评指正，定当努力改正，如有侵权请联系作者删帖。)
)
