200.岛屿数量（bfs写法）

宽搜的正常思路，只不过每次加上计算岛屿数量cnt即可

class Solution {
    public int numIslands(char[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        int dx[] = new int[]{1, -1, 0, 0};
        int dy[] = new int[]{0, 0, 1, -1};
        boolean g[][] = new boolean[n][m];
        Queue<int[]> q = new ArrayDeque<>();
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (g[i][j] == false && grid[i][j] == '1') {
                    q.add(new int[]{i, j});
                    g[i][j] = true;
                    while (!q.isEmpty()) {
                        boolean flag = false;
                        int t[] = q.poll();//弹出队头
                        for (int k = 0; k < 4; k++) {
                            int x = dx[k] + t[0];
                            int y = dy[k] + t[1];
                            if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == '1') {
                                if (g[x][y] == false) {
                                    q.add(new int[]{x, y});
                                    g[x][y] = true;
                                } else {
                                    flag = true;
                                }
                            }
                        }
                        if (flag == false) cnt++;
                    }
                }
            }
        }

        return cnt;
    }
}