dfs查找并集

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。

public class LandSolution {

    private static boolean visited[][];

    public static void main(String[] args) {
        char[][] grid_1 = {
                {'1', '1', '1', '1', '0'},
                {'1', '1', '0', '1', '0'},
                {'1', '1', '0', '0', '0'},
                {'0', '0', '0', '0', '0'}
        };

        char[][] grid_2 = {
                {'1', '1', '0', '0', '0'},
                {'1', '1', '0', '0', '0'},
                {'0', '0', '1', '0', '0'},
                {'0', '0', '0', '1', '1'}
        };

        System.out.println(numIslands(grid_1));
        System.out.println(numIslands(grid_2)); 
    }

    public static int numIslands(char[][] grid) {
        int x = grid.length;
        int y = grid[0].length;
        visited = new boolean[x][y];
        int count = 0;
        for (int i = 0; i < x; i++) {
            for (int j = 0; j < y; j++) {
                if (!visited[i][j] && grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }

        return count;
    }

    private static final int step[][] = {
            {1, 0},
            {-1, 0},
            {0, 1},
            {0, -1}
    };

    public static void dfs(char[][] grid, int i, int j) {
        visited[i][j] = true;
        for (int k = 0; k < step.length; k++) {
            int x = i + step[k][0];
            int y = j + step[k][1];
            if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && grid[x][y] == '1' && !visited[x][y]) {
                dfs(grid, x, y);
            }
        }
    }
}