DFS--岛屿数量 / lake counting

一、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
 

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands

思路：题目翻译过来的意思就是求由“1”组成的极大连通图的个数，所以我们构造dfs函数，dfs函数中对当前陆地“1”和与他邻近的陆地（上下左右四个方向）进行递归处理，将”1“->"0"。这样一次dfs就可以将一个由”1“组成的极大连通图的全部变为"0"，显然我们调用dfs函数的次数便是极大连通图的个数。

class Solution {
    public int numIslands(char[][] grid) {
        in