题目：

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。

方法一：深度优先遍历DFS

class Solution {public int numIslands(char[][] grid) {int count = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == '1') {dfs(grid, i, j);count++;}}}return count;}private void dfs(char[][] grid, int i, int j) {if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') // 终止条件return;grid[i][j] = '0';dfs(grid, i + 1, j); // 下dfs(grid, i, j + 1); // 右dfs(grid, i - 1, j); // 上dfs(grid, i, j - 1); // 左}
}

方法二：广度优先遍历BFS

class Solution {public int numIslands(char[][] grid) {int count = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == '1') {bfs(grid, i, j);count++;}}}return count;}private void bfs(char[][] grid, int i, int j) {Queue<int[]> list = new LinkedList<>();list.add(new int[] {i, j});while (!list.isEmpty()) {int[] cur = list.remove();i = cur[0];j = cur[1];if (i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == '1') {grid[i][j] = '0';list.add(new int[] {i - 1, j});list.add(new int[] {i + 1, j});list.add(new int[] {i, j - 1});list.add(new int[] {i, j + 1});}}}
}