Java 图的遍历-LeetCode200

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

大意就是给个图，找寻所有的子图。注意两点：1.该二维数组并不一定是方阵；2.数组有可能是空的情况

代码如下：

public class Solution {
    class pos {
        public int x;
        public int y;

        public pos(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public int numIslands(char[][] grid) {
        if (grid.length == 0)
            return 0;
        int count = 0;
        int xl = grid.length, yl = grid[0].length;
        int[][] visit = new int[xl][yl];
        LinkedList<pos> lp = new LinkedList<Solution.pos>();
        pos cur;
        for (int i = 0; i < xl; i++) {
            for (int j = 0; j < yl; j++) {
                if (grid[i][j] == '0')
                    continue;
                if (grid[i][j] == '1' && visit[i][j] == 1)
                    continue;
                cur = new pos(i, j);
                lp.push(cur);
                while (lp.size() > 0) {
                    cur = lp.poll();
                    visit[cur.x][cur.y] = 1;
                    if (cur.x + 1 < xl && grid[cur.x + 1][cur.y] == '1'
                            && visit[cur.x + 1][cur.y] != 1) {
                        lp.push(new pos(cur.x + 1, cur.y));
                    }
                    if (cur.x - 1 >= 0 && grid[cur.x - 1][cur.y] == '1'
                            && visit[cur.x - 1][cur.y] != 1) {
                        lp.push(new pos(cur.x - 1, cur.y));
                    }
                    if (cur.y + 1 < yl && grid[cur.x][cur.y + 1] == '1'
                            && visit[cur.x][cur.y + 1] != 1) {
                        lp.push(new pos(cur.x, cur.y + 1));
                    }
                    if (cur.y - 1 >= 0 && grid[cur.x][cur.y - 1] == '1'
                            && visit[cur.x][cur.y - 1] != 1) {
                        lp.push(new pos(cur.x, cur.y - 1));
                    }
                }
                ++count;
            }
        }

        return count;
    }

    public static void main(String[] args) {
        char[][] grid1 = { { '1', '1', '1', '1', '0' },
                { '1', '1', '0', '1', '0' }, { '1', '1', '0', '0', '0' },
                { '0', '0', '0', '0', '0' } }, grid2 = {
                { '1', '1', '0', '0', '0' }, { '1', '1', '0', '0', '0' },
                { '0', '0', '1', '0', '0' }, { '0', '0', '0', '1', '1' } }, grid3 = {
                { '1', '1', '1' }, { '0', '1', '0' }, { '1', '1', '1' } };
        Solution s = new Solution();
        System.out.println(s.numIslands(grid1));
        System.out.println(s.numIslands(grid2));
        System.out.println(s.numIslands(grid3));
    }
}

我利用的是广度优先遍历，也可尝试深度优先遍历