力扣200. 岛屿数量（dfs）

class Solution {
    public int numIslands(char[][] grid) {
       if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int row = grid.length;
        int col = grid[0].length;

        int count = 0;
        //遍历每一个地图上为1的点，对该点进行深度优先搜索
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    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;
        }
        //当前结点已经访问过置成'0'
        grid[i][j] = '0';
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);

    }
    
}