岛屿的个数

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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;    }}

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