给一个01矩阵，求不同的岛屿的个数。

0代表海，1代表岛，如果两个1相邻，那么这两个1属于同一个岛。我们只考虑上下左右为相邻。

class Coordinate {
    int x;
    int y;
    public Coordinate(int x, int y) {
        this.x = x;
        this.y = y;
    }
}
public class Solution {
    /** * @param grid a boolean 2D matrix * @return an integer */
    public int numIslands(boolean[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int n = grid.length;
        int m = grid[0].length;
        int island = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                //搜索到1，island+1,将此岛屿包含的1抹去(标记来过)
                if(grid[i][j]) {
                    markByBFS(grid, i, j);
                    island++;
                }
            }
        }
        return island;
    }
    private void markByBFS(boolean[][] grid, int x, int y) {
        //方向：4个方向
        int[] directionX = {0, 0, -1, 1};
        int[] directionY = {1, -1, 0, 0};
        //BFS的套路，队列，加入根节点，根节点值抹去（标记来过）
        Queue<Coordinate> queue = new LinkedList<>();
        queue.offer(new Coordinate(x, y));
        grid[x][y] = false;
        while (!queue.isEmpty()) {
            Coordinate coor = queue.poll();
            for (int i = 0; i < 4; i++) {
                Coordinate adj = new Coordinate(
                    coor.x + directionX[i],
                    coor.y + directionY[i]
                );
                //判断是否出界
                if (inBound(adj, grid)) {
                   if (grid[adj.x][adj.y]) {
                       grid[adj.x][adj.y] = false;
                       queue.offer(adj);
                   } 
                }
            }
        }
    }
    private boolean inBound(Coordinate coor, boolean[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        return coor.x >= 0 && coor.x < n && coor.y >= 0 && coor.y < m;
    }
}