leecode 200. 岛屿数量

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。

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

public class Solution {

    private int n, m;

    public int numIslands(char[][] grid) {

        n = grid.length;

        if (n == 0) return 0;
        m = grid[0].length;

        int num = 0;

        for (int i = 0; i < n; i++) {

            for (int j = 0; j < m; j++)

                if (grid[i][j] == '1') {
                    DFS(grid, i, j);
                    ++num;
                }
        }
        return num;
    }

    private void DFS(char[][] grid, int i, int j) {

        if (i < 0 || j < 0
                || i >= n || j >= m
                || grid[i][j] != '1')
            return;

        grid[i][j] = '0';

        DFS(grid, i + 1, j);
        DFS(grid, i - 1, j);

        DFS(grid, i, j + 1);
        DFS(grid, i, j - 1);
    }

}